RS Aggarwal Solutions for Class 12 Chapter 20 Homogeneous Differential Equations

Question 1. In each of the following differential equation show that it is homogeneous and solve it. xdy = (x + y)dx
Answer: Starting with xdy = (x + y)dx, we get \( \frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x} = f\left(\frac{y}{x}\right) \)

This shows the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = 1 + \frac{vx}{x} = 1 + v \)

\( x\frac{dv}{dx} = 1 \)

\( dv = \frac{dx}{x} \)

Integrating both sides: \( \int dv = \int \frac{dx}{x} + c \)

\( v = \ln|x| + c \)

Substituting back y = vx: \( \frac{y}{x} = \ln|x| + c \)

\( y = x\ln|x| + cx \)
In simple words: Rearrange the equation to show it depends only on the ratio y/x. Use the substitution y = vx to convert it into variables that separate, integrate both sides, then substitute back to get the final answer.

Exam Tip: Always verify homogeneity by checking that dy/dx can be expressed as a function of y/x alone. After integration, remember to substitute back the original variables.

Question 2. In each of the following differential equation show that it is homogeneous and solve it. (x² - y²)dx + 2xydy = 0
Answer: Starting with (x² - y²)dx + 2xydy = 0, we get \( \frac{dy}{dx} = \frac{-(x^2 - y^2)}{2xy} = \frac{y^2 - x^2}{2xy} \)

This simplifies to \( \frac{dy}{dx} = \frac{y}{2x} - \frac{x}{2y} = \frac{y}{2x} - \frac{1}{2}\left(\frac{x}{y}\right)^{-1} = f\left(\frac{y}{x}\right) \)

This confirms the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{vx}{2x} - \frac{x}{2vx} = \frac{v}{2} - \frac{1}{2v} \)

\( x\frac{dv}{dx} = \frac{v}{2} - \frac{1}{2v} - v = -\frac{v}{2} - \frac{1}{2v} \)

\( x\frac{dv}{dx} = -\frac{v^2 + 1}{2v} \)

\( \frac{2v}{v^2 + 1}dv = -\frac{dx}{x} \)

Integrating both sides: \( \int\frac{2v}{v^2 + 1}dv = -\int\frac{dx}{x} + c \)

\( \ln|v^2 + 1| = -\ln|x| + \ln c \)

Substituting back y = vx: \( \ln\left|\left(\frac{y}{x}\right)^2 + 1\right| + \ln|x| = \ln c \)

\( \left(\left(\frac{y}{x}\right)^2 + 1\right)(x) = c \)

\( x^2 + y^2 = cx \)
In simple words: Express the equation in terms of y/x to confirm homogeneity. Use y = vx as a substitution to separate variables. After integrating and simplifying, the final answer relates x, y, and a constant through a single equation.

Exam Tip: When working with homogeneous equations, carefully track the algebra during substitution and simplification. Always integrate both sides completely before resubstituting.

Question 3. In each of the following differential equation show that it is homogeneous and solve it. x²dy + y(x + y)dx = 0
Answer: Starting with x²dy + y(x + y)dx = 0, we get \( \frac{dy}{dx} = \frac{-y(x + y)}{x^2} = \frac{-(xy + y^2)}{x^2} = -\left(\frac{y}{x} + \frac{y^2}{x^2}\right) = f\left(\frac{y}{x}\right) \)

This shows the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{vx}{2x} - \frac{(vx)^2}{x^2} = \frac{v}{2} - (2v)^{-1} \)

\( v + x\frac{dv}{dx} = \frac{v}{2} - (2v)^{-1} \)

\( x\frac{dv}{dx} = \frac{v}{2} - \frac{1}{2v} - v = -\frac{v}{2} - \frac{1}{2v} \)

\( x\frac{dv}{dx} = -\frac{v^2 + 1}{2v} \)

\( \frac{2v}{v^2 + 1}dv = -\frac{dx}{x} \)

Integrating both sides: \( \int\frac{2v}{v^2 + 1}dv = -\int\frac{dx}{x} + c \)

\( \ln|v^2 + 1| = -\ln|x| + \ln c \)

Substituting back y = vx: \( \ln\left|\left(\frac{y}{x}\right)^2 + 1\right| + \ln|x| = \ln c \)

\( \left(\left(\frac{y}{x}\right)^2 + 1\right)x = c \)

\( x^2 + y^2 = cx \)
In simple words: Rewrite the equation to show it depends only on the ratio y/x. Apply the substitution y = vx and separate the variables. After integrating and substituting back, you obtain a direct relationship between x, y, and the constant.

Exam Tip: Confirm homogeneity by checking whether the derivative can be written as a function of y/x only. Keep careful track of algebraic simplifications during variable separation.

Question 4. In each of the following differential equation show that it is homogeneous and solve it. (x - y)dy - (x + y)dx = 0
Answer: Starting with (x - y)dy - (x + y)dx = 0, we get \( \frac{dy}{dx} = \frac{x + y}{x - y} = \frac{1 + \frac{y}{x}}{1 - \frac{y}{x}} = f\left(\frac{y}{x}\right) \)

This confirms the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = -\left(\frac{vx}{x} + \frac{(vx)^2}{x^2}\right) = -(v + v^2) \)

\( x\frac{dv}{dx} = -v - v^2 - v = -2v - v^2 \)

\( \frac{dv}{2v + v^2} = -\frac{dx}{x} \)

Integrating both sides: \( \int\frac{dv}{2v + v^2} = -\int\frac{dx}{x} + c \)

\( \int\frac{dv}{v(2 + v)} = -\ln|x| + c \)

Using partial fractions: \( \int\frac{1}{2}\ln\left|\frac{v}{2 + v}\right| = -\ln|x| + c \)

\( \ln\left|\frac{v}{2 + v}\right| + 2\ln|x| = 2\ln|c| \)

Substituting back y = vx: \( \ln\left|\frac{y/x}{2 + y/x}\right| + \ln x^2 = \ln|c|^2 \)

\( \ln\left|\frac{y}{y + 2x}\right| + \ln x^2 = \ln|c|^2 \)

\( x^2y = c^2(y + 2x) \)
In simple words: Check that the equation is homogeneous by rewriting it as a function of y/x. Substitute y = vx to separate variables. Integrate both sides carefully, using partial fraction decomposition if needed. Finally, substitute back to obtain the solution.

Exam Tip: When integrating rational functions, use partial fractions to break down complex expressions. Always verify the final answer by differentiating or substituting specific values.

Question 5. In each of the following differential equation show that it is homogeneous and solve it. (x + y)dy + (y - 2x)dx = 0
Answer: Starting with (x + y)dy + (y - 2x)dx = 0, we rearrange to get \( \frac{dy}{dx} = \frac{2x - y}{x + y} = \frac{2 - \frac{y}{x}}{1 + \frac{y}{x}} = f\left(\frac{y}{x}\right) \)

This shows the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{1 + vx/x}{1 - vx/x} = \frac{1 + v}{1 - v} \)

\( x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v(1 - v)}{1 - v} = \frac{1 + v^2}{1 - v} \)

\( \frac{1 - v}{1 + v^2}dv = \frac{dx}{x} \)

Integrating both sides: \( \int\frac{1 - v}{1 + v^2}dv = \int\frac{dx}{x} + c \)

\( \int\frac{1}{1 + v^2}dv - \int\frac{v}{1 + v^2}dv = \ln|x| + c \)

\( \tan^{-1}v - \frac{\ln|1 + v^2|}{2} = \ln|x| + c \)

Substituting back y = vx: \( \tan^{-1}\frac{y}{x} - \frac{\ln\left|1 + \left(\frac{y}{x}\right)^2\right|}{2} = \ln|x| + c \)

\( \tan^{-1}\frac{y}{x} = \frac{\ln|y^2 + x^2|}{2} + \ln|x| + c \)
In simple words: Verify homogeneity by expressing dy/dx as a function of y/x alone. Substitute y = vx and rearrange to separate variables. Integrate using arctangent and logarithm formulas. Substitute back the original variables to find the solution.

Exam Tip: When integrating expressions involving 1 + v², look for opportunities to split the integral into arctangent and logarithmic parts. Keep track of constants throughout the integration process.

Question 6. In each of the following differential equation show that it is homogeneous and solve it. (x² + 3xy + y²)dx - x²dy = 0
Answer: Starting with (x² + 3xy + y²)dx - x²dy = 0, we get \( \frac{dy}{dx} = \frac{x^2 + 3xy + y^2}{x^2} = 1 + 3\frac{y}{x} + \frac{y^2}{x^2} = f\left(\frac{y}{x}\right) \)

This confirms the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = 1 + 3v + v^2 \)

\( x\frac{dv}{dx} = 1 + 2v + v^2 = (1 + v)^2 \)

\( \frac{dv}{(1 + v)^2} = \frac{dx}{x} \)

Integrating both sides: \( \int\frac{dv}{(1 + v)^2} = \int\frac{dx}{x} + c \)

\( -\frac{1}{1 + v} = \ln|x| + c' \)

\( \frac{1}{1 + v} + \ln|x| = c \)

Substituting back y = vx: \( \frac{1}{1 + y/x} + \ln|x| = c \)

\( \frac{x}{x + y} + \ln|x| = c \)
In simple words: Rewrite the equation to show it depends only on y/x. Use the substitution y = vx to convert to separable form. Integrate both sides and simplify. Substitute back to obtain the final solution.

Exam Tip: Recognize perfect squares (1 + v)² in the separated form, which yields simple integration. Always double-check algebraic manipulations when combining fractions.

Question 7. In each of the following differential equation show that it is homogeneous and solve it. 2xydx + (x² + 2y²)dy = 0
Answer: Starting with 2xydx + (x² + 2y²)dy = 0, we get \( \frac{dy}{dx} = -\frac{2xy}{x^2 + 2y^2} = -\frac{2(x/y)}{(x/y)^2 + 2} = f\left(\frac{x}{y}\right) \)

This shows the given differential equation is a homogeneous equation. Note that this is homogeneous in x and y.

To solve, we rearrange: \( \frac{dx}{dy} = -\frac{x^2 + 2y^2}{2xy} \)

Substitute x = vy, so \( \frac{dx}{dy} = v + y\frac{dv}{dy} \)

\( v + y\frac{dv}{dy} = -\frac{(vy)^2 + 2y^2}{2(vy)y} = -\frac{v^2 + 2}{2v} \)

\( y\frac{dv}{dy} = -\frac{v^2 + 2}{2v} - v = -\frac{v^2 + 2 + 2v^2}{2v} = -\frac{3v^2 + 2}{2v} \)

\( \frac{2v}{3v^2 + 2}dv = -\frac{dy}{y} \)

Integrating both sides: \( \int\frac{2v}{3v^2 + 2}dv = -\int\frac{dy}{y} + c \)

\( \frac{\ln|3v^2 + 2|}{3} = -\ln|y| + c' \)

\( \ln|3v^2 + 2| = -3\ln|y| + 3c' \)

Substituting back x = vy: \( \ln\left|3\left(\frac{x}{y}\right)^2 + 2\right| = -3\ln|y| + 3c' \)

\( \ln\left|\frac{3x^2 + 2y^2}{y^2}\right| = -3\ln|y| + 3c' \)

\( 3x^2y + 2y^3 = C \)
In simple words: Recognize that the equation is homogeneous. Since dy/dx is complex, switch to dx/dy and use x = vy. Separate variables and integrate. Substitute back x = vy to find the solution in terms of original variables.

Exam Tip: When working with homogeneous equations, sometimes it is easier to use dx/dy instead of dy/dx. Choose the approach that leads to simpler integration.

Question 8. In each of the following differential equation show that it is homogeneous and solve it. \( \frac{dy}{dx} + \frac{x - 2y}{2x - y} = 0 \)
Answer: Starting with \( \frac{dy}{dx} + \frac{x - 2y}{2x - y} = 0 \), we get \( \frac{dy}{dx} = \frac{2y - x}{2x - y} = \frac{2(y/x) - 1}{2 - (y/x)} = f\left(\frac{y}{x}\right) \)

This confirms the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{2v - 1}{2 - v} \)

\( x\frac{dv}{dx} = \frac{2v - 1}{2 - v} - v = \frac{2v - 1 - v(2 - v)}{2 - v} = \frac{v^2 - 1}{2 - v} \)

\( \frac{2 - v}{v^2 - 1}dv = \frac{dx}{x} \)

\( \frac{v - 2}{v^2 - 1}dv = -\frac{dx}{x} \)

\( \frac{v}{v^2 - 1}dv - \frac{2}{v^2 - 1}dv = -\frac{dx}{x} \)

Using partial fractions, integrate both sides: \( \frac{\ln|v^2 - 1|}{2} - 2 \times \frac{1}{2}\ln\left|\frac{v - 1}{v + 1}\right| = -\ln|x| + c \)

\( \frac{\ln|v^2 - 1|}{2} - \ln\left|\frac{v - 1}{v + 1}\right| = -\ln|x| + c \)

Substituting back y = vx: \( (y - x) = C(y + x)^3 \)
In simple words: Verify homogeneity by confirming the derivative can be expressed as a function of y/x. Apply y = vx and simplify. Use partial fraction decomposition during integration. Substitute back to get the final answer.

Exam Tip: When dealing with v² - 1 = (v - 1)(v + 1), use partial fractions to simplify integration. Track logarithmic terms carefully during combination and substitution.

Question 9. In each of the following differential equation show that it is homogeneous and solve it. \( \frac{dy}{dx} + \frac{x^2 - y^2}{3xy} = 0 \)
Answer: Starting with \( \frac{dy}{dx} + \frac{x^2 - y^2}{3xy} = 0 \), we get \( \frac{dy}{dx} = \frac{y^2 - x^2}{3xy} = \frac{(y/x)^2 - 1}{3(y/x)} = f\left(\frac{y}{x}\right) \)

This shows the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{v^2 - 1}{3v} \)

\( x\frac{dv}{dx} = \frac{v^2 - 1}{3v} - v = \frac{v^2 - 1 - 3v^2}{3v} = \frac{-2v^2 - 1}{3v} \)

\( x\frac{dv}{dx} = -\frac{2v^2 + 1}{3v} \)

\( \frac{3v}{2v^2 + 1}dv = -\frac{dx}{x} \)

Integrating both sides: \( \frac{3}{2}\int\frac{2v}{2v^2 + 1}dv = -\int\frac{dx}{x} + c \)

\( \frac{3}{2}\ln|2v^2 + 1| = -\ln|x| + c' \)

\( \ln|2v^2 + 1| = -\frac{2}{3}\ln|x| + c'' \)

Substituting back y = vx: \( \ln\left|2\left(\frac{y}{x}\right)^2 + 1\right| = -\frac{2}{3}\ln|x| + c'' \)

\( (x^2 + 2y^2)^3 = Cx^2 \)
In simple words: Check homogeneity by expressing dy/dx as a function of y/x. Use the substitution y = vx to achieve variable separation. Integrate using standard logarithm techniques. Substitute back to obtain the solution.

Exam Tip: When integrating terms like \( \frac{2v}{2v^2 + 1} \), recognize this as the derivative of ln|2v² + 1|. Pay attention to constant factors throughout the integration.

Question 10. In each of the following differential equation show that it is homogeneous and solve it. \( \frac{dy}{dx} + \frac{x^2 - y^2}{3xy} = 0 \)
Answer: Starting with \( \frac{dy}{dx} = -\frac{x^2 - y^2}{3xy} \), we get \( \frac{dy}{dx} = -\left(\frac{y}{3x} - \frac{x}{3y}\right) = -\left(\frac{(y/x)^{-1}}{3} + \frac{1}{3(y/x)}\right) = f\left(\frac{y}{x}\right) \)

This confirms the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = -\left(\frac{vx}{3x}\right)^{-1} + \frac{1}{3(vx/x)} = -\left(\frac{v}{3}\right)^{-1} + \frac{1}{3v} \)

\( x\frac{dv}{dx} = -\frac{3}{v} + \frac{1}{3v} = \frac{-9 + v^2}{3v} \)

\( \frac{3v}{v^2 - 9}dv = \frac{dx}{x} \)

Integrating both sides: \( \int\frac{3v}{v^2 - 9}dv = \int\frac{dx}{x} + c \)

\( \frac{3}{2}\ln|v^2 - 9| = \ln|x| + c' \)

Substituting back y = vx: \( \frac{3}{2}\ln\left|\left(\frac{y}{x}\right)^2 - 9\right| = \ln|x| + c' \)

\( (x^2 + 2y^2)^3 = Cx^2 \)
In simple words: Verify the equation is homogeneous. Apply y = vx to convert to separable form. Integrate both sides using the logarithm formula for rational functions. Substitute back to find the solution.

Exam Tip: When integrating v/(v² - 9), factor the denominator and use logarithm properties. Keep exponents and coefficients organized throughout calculations.

Question 11. In each of the following differential equation show that it is homogeneous and solve it. \( \frac{dy}{dx} = \frac{2xy}{x^2 - y^2} \)
Answer: Starting with \( \frac{dy}{dx} = \frac{2xy}{x^2 - y^2} \), we get \( \frac{dy}{dx} = \frac{2(y/x)}{1 - (y/x)^2} = f\left(\frac{y}{x}\right) \)

This confirms the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{2v}{1 - v^2} \)

\( x\frac{dv}{dx} = \frac{2v}{1 - v^2} - v = \frac{2v - v(1 - v^2)}{1 - v^2} = \frac{2v - v + v^3}{1 - v^2} = \frac{v + v^3}{1 - v^2} \)

\( \frac{1 - v^2}{v(1 + v^2)}dv = \frac{dx}{x} \)

Integrating both sides: \( \int\frac{1 - v^2}{v(1 + v^2)}dv = \int\frac{dx}{x} + c \)

\( \ln|v| - \frac{1}{2}\ln|1 + v^2| = \ln|x| + c' \)

\( \ln\left|\frac{v}{\sqrt{1 + v^2}}\right| = \ln|x| + c' \)

Substituting back y = vx: \( y = C(y^2 + x^2) \)
In simple words: Establish homogeneity by showing dy/dx depends only on y/x. Substitute y = vx to separate variables. Use partial fractions if needed during integration. Substitute back to obtain the final solution.

Exam Tip: When integrating complex rational expressions, decompose using partial fractions. Recognize that ln|1 + v²| simplifies when combined with other logarithmic terms.

Question 12. In each of the following differential equation show that it is homogeneous and solve it. \( \frac{dy}{dx} = \frac{2xy}{x^2 - y^2} \)
Answer: Starting with \( \frac{dy}{dx} = \frac{2xy}{x^2 - y^2} \), we get \( \frac{dy}{dx} = \frac{2(y/x)}{1 - (y/x)^2} = f\left(\frac{y}{x}\right) \)

This shows the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{2v}{1 - v^2} \)

\( x\frac{dv}{dx} = \frac{2v - v(1 - v^2)}{1 - v^2} = \frac{v + v^3}{1 - v^2} \)

\( \frac{(1 - v^2)}{v(1 + v^2)}dv = \frac{dx}{x} \)

Integrating both sides: \( \int\frac{(1 - v^2)}{v(1 + v^2)}dv = \int\frac{dx}{x} + c \)

\( \ln|(v)^2 - 1| = \ln|x| + c' \)

Substituting back y = vx: \( \ln\left|\left(\frac{y}{x}\right)^2 - 1\right| = \ln|x| + c' \)

\( y = C(y^2 + x^2) \)
In simple words: Check that the equation is homogeneous in terms of y/x. Use the substitution y = vx and carefully perform algebra to separate variables. Integrate both sides and substitute back to find the solution.

Exam Tip: Always verify that after substitution and simplification, variables are truly separated (one side has only v and dv, the other only x and dx). This ensures integration is valid.

Question 13. In each of the following differential equation show that it is homogeneous and solve it. \( x^2\frac{dy}{dx} = x^2 + xy + y^2 \)
Answer: Starting with \( x^2\frac{dy}{dx} = x^2 + xy + y^2 \), we get \( \frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 = f\left(\frac{y}{x}\right) \)

This confirms the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = 1 + v + v^2 \)

\( x\frac{dv}{dx} = 1 + v^2 \)

\( \frac{dv}{1 + v^2} = \frac{dx}{x} \)

Integrating both sides: \( \int\frac{dv}{1 + v^2} = \int\frac{dx}{x} + c \)

\( \tan^{-1}v = \ln|x| + c' \)

Substituting back y = vx: \( \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + c' \)
In simple words: Establish homogeneity by expressing dy/dx as a function of y/x only. Apply y = vx and separate variables into form 1/(1 + v²) dv = dx/x. Integrate both sides and substitute back to the original variables.

Exam Tip: Recognize that the integral of 1/(1 + v²) is tan⁻¹(v). Make sure to include the constant of integration and handle it correctly during substitution.

Question 14. In each of the following differential equation show that it is homogeneous and solve it. \( y^2 + (x^2 - xy)\frac{dy}{dx} = 0 \)
Answer: Starting with \( y^2 + (x^2 - xy)\frac{dy}{dx} = 0 \), we rearrange: \( \frac{dy}{dx} = \frac{-y^2}{x^2 - xy} \)

Rewriting: \( \frac{dx}{dy} = \frac{x^2 - xy}{-y^2} = -\frac{x^2}{y^2} + \frac{x}{y} = -\left(\frac{x}{y}\right)^{-1} + \frac{x}{y} = f\left(\frac{x}{y}\right) \)

This shows the given differential equation is a homogeneous equation.

To solve, substitute x = vy, so \( \frac{dx}{dy} = v + y\frac{dv}{dy} \)

\( v + y\frac{dv}{dy} = \frac{vy - v^2y}{y^2} = \frac{v - v^2}{y} \)

\( y\frac{dv}{dy} = \frac{v - v^2 - vy}{y} = \frac{v - v^2 - v}{y} = \frac{-v^2}{y} \)

\( \frac{dv}{v^2} = -\frac{dy}{y} \)

Integrating both sides: \( \int\frac{dv}{v^2} = -\int\frac{dy}{y} + c \)

\( -\frac{1}{v} = -\ln|y| + c' \)

\( \frac{1}{v} + \ln|y| = c \)

Substituting back x = vy: \( \frac{y}{x} + \ln|y| = c \)

\( y = x(\ln|y| + c) \)
In simple words: Note that this equation involves dy/dx in a form that is easier to work with as dx/dy. Check homogeneity. Apply the substitution x = vy. Separate variables and integrate using standard formulas. Substitute back to obtain the solution.

Exam Tip: When dy/dx leads to complicated algebra, try switching to dx/dy and using x = vy instead. This often simplifies the solution process significantly.

Question 15. In each of the following differential equation show that it is homogeneous and solve it. \( x\frac{dy}{dx} - y = 2\sqrt{y^2 - x^2} \)
Answer: Starting with \( x\frac{dy}{dx} - y = 2\sqrt{y^2 - x^2} \), we rearrange: \( x\frac{dy}{dx} = y + 2\sqrt{y^2 - x^2} \)

\( \frac{dy}{dx} = \frac{y + 2\sqrt{y^2 - x^2}}{x} = \frac{y}{x} + 2\sqrt{\left(\frac{y}{x}\right)^2 - 1} = f\left(\frac{y}{x}\right) \)

This confirms the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = v + 2\sqrt{v^2 - 1} \)

\( x\frac{dv}{dx} = 2\sqrt{v^2 - 1} \)

\( \frac{dv}{\sqrt{v^2 - 1}} = 2\frac{dx}{x} \)

Integrating both sides: \( \int\frac{dv}{\sqrt{v^2 - 1}} = 2\int\frac{dx}{x} + c \)

\( \ln|v + \sqrt{v^2 - 1}| = 2\ln|x| + c' \)

Substituting back y = vx: \( \ln\left|\frac{y}{x} + \sqrt{\left(\frac{y}{x}\right)^2 - 1}\right| = 2\ln|x| + c' \)

\( y + \sqrt{y^2 - x^2} = C|x|^3 \)
In simple words: Establish homogeneity by showing dy/dx is a function of y/x. Substitute y = vx and simplify. Recognize the form 1/√(v² - 1) which integrates to ln|v + √(v² - 1)|. Substitute back to find the solution.

Exam Tip: Know the standard integral of 1/√(v² - 1), which equals ln|v + √(v² - 1)| or sinh⁻¹(v). This appears frequently in homogeneous equations involving square roots.

Question 16. In each of the following differential equation show that it is homogeneous and solve it. \( y^2dx + (x^2 + xy + y^2)dy = 0 \)
Answer: Starting with \( y^2dx + (x^2 + xy + y^2)dy = 0 \), we rearrange: \( \frac{dx}{dy} = -\frac{x^2 + xy + y^2}{y^2} = -\left(1 + \frac{x}{y} + \left(\frac{x}{y}\right)^2\right) = f\left(\frac{x}{y}\right) \)

This shows the given differential equation is a homogeneous equation.

To solve, substitute x = vy, so \( \frac{dx}{dy} = v + y\frac{dv}{dy} \)

\( v + y\frac{dv}{dy} = -(1 + v + v^2) \)

\( y\frac{dv}{dy} = -1 - v - v^2 - v = -(1 + 2v + v^2) = -(1 + v)^2 \)

\( \frac{dv}{(1 + v)^2} = -\frac{dy}{y} \)

Integrating both sides: \( \int\frac{dv}{(1 + v)^2} = -\int\frac{dy}{y} + c \)

\( -\frac{1}{1 + v} = -\ln|y| + c' \)

\( \frac{1}{1 + v} + \ln|y| = c \)

Substituting back x = vy: \( \frac{y}{x + y} + \ln|y| = c \)

\( \log\left|\frac{y}{x + y}\right| + \log|x| + \frac{x}{(y + x)} = C \)
In simple words: Verify homogeneity by expressing dx/dy as a function of x/y. Use x = vy to convert to separable form. Integrate both sides carefully. Substitute back to the original variables to find the solution.

Exam Tip: Recognize perfect squares like (1 + v)² during variable separation, which simplifies integration. Always verify the final answer makes sense by checking limiting cases.

Question 17. In each of the following differential equation show that it is homogeneous and solve it. \( (x - y)\frac{dy}{dx} = x + 3y \)
Answer: Starting with \( (x - y)\frac{dy}{dx} = x + 3y \), we rearrange: \( \frac{dy}{dx} = \frac{x + 3y}{x - y} = \frac{1 + 3(y/x)}{1 - (y/x)} = f\left(\frac{y}{x}\right) \)

This confirms the given differential equation is a homogeneous equation.

To solve, substitute y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{1 + 3v}{1 - v} \)

\( x\frac{dv}{dx} = \frac{1 + 3v - v(1 - v)}{1 - v} = \frac{1 + 3v - v + v^2}{1 - v} = \frac{1 + 2v + v^2}{1 - v} = \frac{(1 + v)^2}{1 - v} \)

But rearranging further: \( x\frac{dv}{dx} = \frac{1 + 2v + v^2 + v^3}{1 - v} \)

\( \frac{(1 - v)dv}{1 + v + v^2 + v^3} = \frac{dy}{y} \)

Integrating both sides: \( \int\frac{(1 - v)dv}{1 + v + v^2 + v^3} = \int\frac{dy}{y} + c \)

Using partial fractions and careful integration, this evaluates to: \( \log\left|\frac{y}{y + x}\right| + \log|x| + \frac{x}{(y + x)} = C \)
In simple words: Establish homogeneity by showing dy/dx depends only on y/x. Substitute y = vx to separate variables. Use partial fraction decomposition where necessary. Integrate and substitute back to obtain the final solution.

Exam Tip: For equations that require partial fraction decomposition, set up the decomposition carefully and verify by expanding. Keep track of all logarithmic and rational terms during integration.

Question 18. In each of the following differential equation show that it is homogeneous and solve it. (x³ + 3xy²)dx + (y³ + 3x²y)dy = 0
Answer: Rewriting the equation:
\( (x^3 + 3xy^2)dx + (y^3 + 3x^2y)dy = 0 \)

Finding dy/dx:
\( \frac{dy}{dx} = -\frac{x^3 + 3xy^2}{y^3 + 3x^2y} = -\frac{3xy^2}{3x^2y} \left(\frac{\frac{x^2}{3xy^2} + 1}{\frac{y^2}{3x^2y} + 1}\right) = -\frac{y}{x} \left(\frac{\left(\frac{x}{y}\right)^2 + 1}{\left(\frac{y}{x}\right)^2 + 1}\right) \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

This demonstrates the differential equation is homogeneous in nature.

To find the solution, apply the substitution y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

After substitution and simplification through integration, the general solution is:
\( y^4 + 6x^2y^2 + x^4 = C \)
In simple words: This differential equation is homogeneous because the right side depends only on the ratio y/x. Using the standard substitution technique and integration gives us the solution as shown.

Exam Tip: Always verify homogeneity by checking if dy/dx can be written as a function of y/x alone. Use the substitution y = vx systematically and integrate carefully.

Question 19. In each of the following differential equation show that it is homogeneous and solve it. (x - \(\sqrt{xy}\))dy = ydx
Answer: Starting with the equation:
\( \frac{dy}{dx} = \frac{y}{x - \sqrt{xy}} = \frac{1}{\frac{x}{y} - \sqrt{\frac{x}{y}}} = \frac{1}{\left(\frac{y}{x}\right)^{-1} - \sqrt{\left(\frac{y}{x}\right)^{-1}}} \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To obtain the solution, substitute y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{1}{\left(\frac{vx}{x}\right)^{-1} - \sqrt{\left(\frac{vx}{x}\right)^{-1}}} = \frac{1}{v^{-1} - \frac{1}{\sqrt{v}}} = \frac{v\sqrt{v}}{\sqrt{v} - v} \)

\( x\frac{dv}{dx} = \frac{v\sqrt{v}}{\sqrt{v} - v} - v = \frac{v\sqrt{v} - v\sqrt{v} + v^2}{\sqrt{v} - v} = \frac{v^2}{\sqrt{v} - v} \)

\( \frac{\sqrt{v} - v}{v^2}dv = \frac{dx}{x} \)

\( \frac{1}{v\sqrt{v}}dv - \frac{1}{v}dv = \frac{dx}{x} \)

Integrating both sides:
\( \int\frac{1}{v\sqrt{v}}dv - \int\frac{1}{v}dv = \int\frac{dx}{x} + c \)

\( -\frac{1}{\sqrt{v}} - \ln|v| = \ln|x| + c \)

Resubstituting y = vx:
\( -\frac{1}{\sqrt{\frac{y}{x}}} - \ln\left|\frac{y}{x}\right| = \ln|x| + c \)

\( 2\sqrt{\frac{x}{y}} + \log|y| = C \)

\( y^4 + 6x^2y^2 + x^4 = C \)
In simple words: This equation is homogeneous since dy/dx can be written as a function of y/x. The substitution y = vx transforms it into a separable form that can be integrated.

Exam Tip: When dealing with square roots in homogeneous equations, carefully handle the algebra during separation of variables before integrating.

Question 20. In each of the following differential equation show that it is homogeneous and solve it. x² dy/dx + y² = xy
Answer: Rearranging the equation:
\( \frac{dy}{dx} = \frac{xy - y^2}{x^2} = \frac{y}{x} - \left(\frac{y}{x}\right)^2 \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To find the solution, let y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{vx}{x} - \left(\frac{vx}{x}\right)^2 = v - v^2 \)

\( x\frac{dv}{dx} = -v^2 \)

\( \frac{dv}{-v^2} = \frac{dx}{x} \)

Integrating both sides:
\( \int\frac{dv}{-v^2} = \int\frac{dx}{x} + c \)

\( \frac{1}{v} = \ln|x| + c \)

Resubstituting y = vx:
\( \frac{1}{\frac{y}{x}} = \ln|x| + c \)

\( \frac{x}{y} = \ln|x| + \ln|c| \)

\( \frac{x}{y} = \ln|xc| \)
In simple words: The equation is homogeneous because dy/dx depends only on the ratio y/x. After substituting y = vx and separating variables, integration gives the relationship between x and y.

Exam Tip: Separate variables carefully and ensure all constant terms are combined into a single constant C for the final answer.

Question 21. In each of the following differential equation show that it is homogeneous and solve it. x dy/dx = y(log y - log x + 1)
Answer: Starting with:
\( x\frac{dy}{dx} = y\left(\log\frac{y}{x} + 1\right) \)

\( \frac{dy}{dx} = \frac{y}{x}\left(\log\frac{y}{x} + 1\right) \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To solve, substitute y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{vx}{x}\left(\log\frac{vx}{x} + 1\right) = v(\log v + 1) \)

\( x\frac{dv}{dx} = v\log v \)

\( \frac{dv}{v\log v} = \frac{dx}{x} \)

Integrating both sides:
\( \int\frac{dv}{v\log v} = \int\frac{dx}{x} + c \)

\( \log|\log v| = \log|xc| \)

\( \log|v| = xc \)

\( v = e^{xc} \)

Resubstituting y = vx:
\( y = xe^{xc} \)
In simple words: This equation is homogeneous because the right side depends only on y/x. Use the substitution y = vx, separate variables, and integrate to get the final solution.

Exam Tip: Watch for logarithmic terms - the integral of 1/(v log v) requires recognizing it as a standard form. Always verify your final answer by differentiation if time permits.

Question 22. In each of the following differential equation show that it is homogeneous and solve it. x dy/dx - y + x sin(y/x) = 0
Answer: Rewriting the equation:
\( x\frac{dy}{dx} = y - x\sin\frac{y}{x} \)

\( \frac{dy}{dx} = \frac{y}{x} - \sin\frac{y}{x} \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To find the solution, let y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{vx}{x} - \sin\frac{vx}{x} = v - \sin v \)

\( x\frac{dv}{dx} = -\sin v \)

\( \frac{dv}{\sin v} = -\frac{dx}{x} \)

Integrating both sides:
\( \int\frac{dv}{\sin v} = -\int\frac{dx}{x} + c \)

\( \log\tan\left(\frac{v}{2}\right) = -\log|x| + c \)

Resubstituting y = vx:
\( \log\tan\left(\frac{y}{2x}\right) = -\log|x| + \log c \)

\( x\tan\left(\frac{y}{2x}\right) = C \)
In simple words: This is a homogeneous equation since the derivative can be expressed as a function of y/x only. After substitution and integration, we obtain the relationship shown above.

Exam Tip: The integral of 1/sin v equals log|tan(v/2)|. Remember this standard form when integrating trigonometric reciprocals.

Question 23. In each of the following differential equation show that it is homogeneous and solve it. x dy/dx = y - x cos²(y/x)
Answer: Starting with:
\( x\frac{dy}{dx} = y - x\cos^2\frac{y}{x} \)

\( \frac{dy}{dx} = \frac{y}{x} - \cos^2\frac{y}{x} \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To obtain the solution, substitute y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{vx}{x} - \cos^2\frac{vx}{x} = v - \cos^2 v \)

\( x\frac{dv}{dx} = -\cos^2 v \)

\( \frac{dv}{\cos^2 v} = -\frac{dx}{x} \)

Integrating both sides:
\( \int\frac{dv}{\cos^2 v} = -\int\frac{dx}{x} + c \)

\( \tan v = -\ln|x| + c \)

Resubstituting y = vx:
\( \tan\left(\frac{y}{x}\right) + \ln|x| = c \)
In simple words: This is a homogeneous differential equation where the right side is a function of y/x. The substitution y = vx makes it separable, and integration yields the final form.

Exam Tip: The integral of sec²v equals tan v. Always include the logarithmic term when combining constants from both sides of the integration.

Question 24. In each of the following differential equation show that it is homogeneous and solve it. (x cos(y/x)) dy/dx = (y cos(y/x)) + x
Answer: Rewriting:
\( \left(x\cos\frac{y}{x}\right)\frac{dy}{dx} = y\cos\frac{y}{x} + x \)

\( \frac{dy}{dx} = \frac{y\cos\frac{y}{x} + x}{x\cos\frac{y}{x}} \)

\( \frac{dy}{dx} = \frac{y}{x} + \sec\frac{y}{x} \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To find the solution, let y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = v + \sec v \)

\( x\frac{dv}{dx} = \sec v \)

\( \frac{dv}{\sec v} = \frac{dx}{x} \)

\( \cos v \, dv = \frac{dx}{x} \)

Integrating both sides:
\( \int\cos v \, dv = \int\frac{dx}{x} + c \)

\( \sin v = \ln|x| + c \)

Resubstituting y = vx:
\( \sin\left(\frac{y}{x}\right) = \ln|x| + c \)
In simple words: The equation is homogeneous since dy/dx can be written as a function of y/x only. After substitution and separation of variables, integration produces the solution shown.

Exam Tip: Watch for cancellation opportunities after substitution - here the cosine terms simplify nicely, making the equation separable.

Question 25. Find the particular solution of the differential equation. 2xy + y² - 2x² dy/dx = 0, it being given that y = 2 when x = 1
Answer: Starting with:
\( 2xy + y^2 - 2x^2\frac{dy}{dx} = 0 \)

\( \frac{dy}{dx} = \frac{2xy + y^2}{2x^2} = \frac{y}{x} + \frac{y^2}{2x^2} \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To find the general solution, substitute y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{vx}{x} + \frac{(vx)^2}{2x^2} = v + \frac{v^2}{2} \)

\( x\frac{dv}{dx} = \frac{v^2}{2} \)

\( \frac{dv}{v^2} = \frac{2dx}{x} \)

Integrating both sides:
\( \int\frac{dv}{v^2} = 2\int\frac{dx}{x} + c \)

\( -\frac{1}{v} = 2\ln|x| + c \)

Resubstituting y = vx:
\( -\frac{x}{y} = 2\ln|x| + c \)

Now apply the initial condition y = 2 when x = 1:
\( -\frac{1}{2} = 2\ln(1) + c \)
\( -\frac{1}{2} = 0 + c \)
\( c = -\frac{1}{2} \)

Therefore:
\( -\frac{x}{y} = 2\ln|x| - \frac{1}{2} \)

\( y = \frac{2x}{1 - 4\ln|x|} \)
In simple words: After recognizing this as a homogeneous equation, use the substitution y = vx to make it separable. Integrate and apply the boundary condition to find the particular solution.

Exam Tip: Always verify that initial conditions are correctly substituted to find the constant C. Check your final answer by substituting the given point back into it.

Question 26. Find the particular solution of the differential equation. (x sin²(y/x) - y)dx + xdy = 0, it being given that y = π/4 when x = 1
Answer: Rearranging:
\( \frac{dy}{dx} = \frac{y - x\sin^2\left(\frac{y}{x}\right)}{x} = \frac{y}{x} - \sin^2\left(\frac{y}{x}\right) \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To obtain the general solution, let y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{vx}{x} - \sin^2\left(\frac{vx}{x}\right) = v - \sin^2 v \)

\( x\frac{dv}{dx} = -\sin^2 v \)

\( \frac{dv}{\sin^2 v} = -\frac{dx}{x} \)

Integrating both sides:
\( \int\frac{dv}{\sin^2 v} = -\int\frac{dx}{x} + c \)

\( \cot v = \ln|x| + c \)

Resubstituting y = vx:
\( \cot\left(\frac{y}{x}\right) = \ln|x| + c \)

Apply the initial condition y = π/4 when x = 1:
\( \cot\left(\frac{\pi/4}{1}\right) = \ln(1) + c \)
\( \cot\left(\frac{\pi}{4}\right) = 0 + c \)
\( 1 = c \)

Therefore:
\( \cot\left(\frac{y}{x}\right) = \ln|x| + 1 \)
In simple words: This is a homogeneous equation where the derivative depends only on y/x. Substitute y = vx to separate variables, integrate, then use the given boundary values to determine the constant.

Exam Tip: Remember that the integral of cosec²v equals -cot v. Always double-check that your initial condition values are exact (like cot(π/4) = 1) before proceeding.

Question 27. Find the particular solution of the differential equation dy/dx = y(2y - x) / (x(2y + x)), given that y = 1 when x = 1.
Answer: Beginning with:
\( \frac{dy}{dx} = \frac{y(2y - x)}{x(2y + x)} \)

\( \frac{dy}{dx} = \frac{y\left(2\frac{y}{x} - 1\right)}{x\left(2\frac{y}{x} + 1\right)} \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To find the general solution, substitute y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{vx(2v - 1)}{x(2v + 1)} = v\left(\frac{2v - 1}{2v + 1}\right) \)

\( x\frac{dv}{dx} = v\left(\frac{2v - 1}{2v + 1}\right) - v = v\left(\frac{2v - 1 - 2v - 1}{2v + 1}\right) = -\frac{2v}{2v + 1} \)

\( \frac{2v + 1}{2v}dv = -\frac{dx}{x} \)

\( dv + \frac{1}{2v}dv = -\frac{dx}{x} \)

Integrating both sides:
\( \int\left(dv + \frac{1}{2v}dv\right) = -\int\frac{dx}{x} + c \)

\( v + \frac{\ln|v|}{2} = -\ln|x| + c \)

Resubstituting y = vx:
\( \frac{y}{x} + \frac{\ln\left|\frac{y}{x}\right|}{2} = -\ln|x| + c \)

Apply the initial condition y = 1 when x = 1:
\( 1 + \frac{\ln|1|}{2} = -\ln|1| + c \)
\( 1 + 0 = 0 + c \)
\( c = 1 \)

Therefore:
\( \frac{y}{x} + \frac{\ln\left|\frac{y}{x}\right|}{2} = -\ln|x| + 1 \)
In simple words: Identify this as a homogeneous equation by factoring out x terms. Use y = vx to transform it into a separable form, then integrate term by term and apply the boundary condition.

Exam Tip: When a rational function can be decomposed, split the integral accordingly. Always verify your logarithmic manipulations match the given boundary values.

Question 28. Find the particular solution of the differential equation xe^(y/x) - y + x dy/dx = 0, given that y(1) = 0.
Answer: Rearranging:
\( x\frac{dy}{dx} = y - xe^{y/x} \)

\( \frac{dy}{dx} = \frac{y}{x} - e^{y/x} \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To obtain the general solution, substitute y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{vx}{x} - e^{vx/x} = v - e^v \)

\( x\frac{dv}{dx} = -e^v \)

\( \frac{dv}{e^v} = -\frac{dx}{x} \)

Integrating both sides:
\( \int\frac{dv}{e^v} = -\int\frac{dx}{x} + c \)

\( -e^{-v} = -\ln|x| + c \)

Resubstituting y = vx:
\( -e^{-y/x} = -\ln|x| + c \)

Apply the initial condition y(1) = 0:
\( -e^{0} = -\ln(1) + c \)
\( -1 = 0 + c \)
\( c = -1 \)

Therefore:
\( -e^{-y/x} = -\ln|x| - 1 \)

\( \log|x| + e^{-y/x} = 1 \)
In simple words: This is homogeneous because dy/dx can be expressed in terms of y/x. The substitution y = vx makes it separable. Integration and the initial condition yield the particular solution.

Exam Tip: Exponential integrals like ∫e^(-v)dv need careful attention to signs. Always recheck your constant by substituting the initial condition back into the general solution.

Question 29. Find the particular solution of the differential equation xe^(y/x) - y + x dy/dx = 0, given that y(e) = 0.
Answer: Starting from:
\( x\frac{dy}{dx} = y - xe^{y/x} \)

\( \frac{dy}{dx} = \frac{y}{x} - e^{y/x} \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To find the general solution, let y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = v - e^v \)

\( x\frac{dv}{dx} = -e^v \)

\( \frac{dv}{e^v} = -\frac{dx}{x} \)

Integrating both sides:
\( -e^{-v} = -\ln|x| + c \)

Resubstituting y = vx:
\( -e^{-y/x} = -\ln|x| + c \)

Apply the initial condition y(e) = 0:
\( -e^{0} = -\ln(e) + c \)
\( -1 = -1 + c \)
\( c = 0 \)

Therefore:
\( -e^{-y/x} = -\ln|x| \)

\( y = -x\log(\log|x|) \)
In simple words: Recognize this as a homogeneous equation since the derivative depends only on y/x. The substitution y = vx makes it separable, and the initial condition at (e, 0) determines the constant.

Exam Tip: When the initial condition is at a special point like x = e, the logarithmic terms simplify nicely - always evaluate these carefully.

Question 30. The slope of the tangent to a curve at any point (x,y) on it is given by \( \frac{y}{x} - \cot\frac{y}{x}\cos\frac{y}{x} \), where x > 0 and y > 0. If the curve passes through the point \( \left(1, \frac{\pi}{4}\right) \), find the equation of the curve.
Answer: The slope of the tangent is the derivative, so:
\( \frac{dy}{dx} = \frac{y}{x} - \cot\frac{y}{x}\cos\frac{y}{x} \)

\( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \)

The given differential equation is homogeneous.

To find the solution, substitute y = vx:
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

\( v + x\frac{dv}{dx} = \frac{vx}{x} - \cot\frac{vx}{x}\cos\frac{vx}{x} = v - \cot v\cos v \)

\( x\frac{dv}{dx} = -\cot v\cos v \)

\( \frac{dv}{-\cot v\cos v} = \frac{dx}{x} \)

Integrating both sides:
\( \int\frac{dv}{-\cot v\cos v} = \int\frac{dx}{x} + c \)

\( -\frac{1}{\cos v} = \ln|x| + c \)

Resubstituting y = vx:
\( -\frac{1}{\cos\frac{y}{x}} = \ln|x| + c \)

Apply the condition that the curve passes through \( \left(1, \frac{\pi}{4}\right) \):
\( -\frac{1}{\cos\frac{\pi/4}{1}} = \ln(1) + c \)
\( -\frac{1}{\cos\frac{\pi}{4}} = 0 + c \)
\( -\frac{1}{\frac{1}{\sqrt{2}}} = c \)
\( c = -\sqrt{2} \)

Therefore:
\( \sec\frac{y}{x} + \log|x| = \sqrt{2} \)
In simple words: This is a homogeneous differential equation where the slope depends only on the ratio y/x. Using the standard substitution y = vx converts it to a separable form. The boundary condition at the given point determines the constant.

Exam Tip: When integrating expressions involving cosines and cotangents, recognize -1/cos(v) = -sec(v). Always compute trigonometric values at special angles like π/4 correctly - this is where many errors occur.