Question

The general solution of the differential equation $(x^3-y^3)dx = (x^2y-xy^2)dy$ is

• A. $y=x\log(c|x+y|)$
• B. $y=\log(c|x+y|)$
• C. $xy = \log(c|x+y|)$
• D. $x+y+\log|x+y|+c=0$

Hint:

To quickly check if a differential equation of the form $M(x,y)dx+N(x,y)dy=0$ is homogeneous, check if $M(tx,ty)=t^k M(x,y)$ and $N(tx,ty)=t^k N(x,y)$ for the same degree $k$. If it is, the substitution $y=vx$ (or $x=vy$) is the standard method to solve it.

Answer 1

The Correct Option is A

Step 1: Simplify the Differential Equation The given equation is \( (x^3 - y^3)dx = xy(x - y)dy \). Factorize the LHS using \( x^3 - y^3 = (x-y)(x^2 + xy + y^2) \): \[ (x-y)(x^2 + xy + y^2)dx = xy(x - y)dy \] Assuming \( x \neq y \), we can cancel \( (x-y) \): \[ (x^2 + xy + y^2)dx = xy dy \] \[ \frac{dy}{dx} = \frac{x^2 + xy + y^2}{xy} \] \[ \frac{dy}{dx} = \frac{x}{y} + 1 + \frac{y}{x} \]
Step 2: Solve the Homogeneous Equation This is a homogeneous differential equation. Put \( y = vx \), so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substitute into the equation: \[ v + x\frac{dv}{dx} = \frac{1}{v} + 1 + v \] Cancel \( v \) from both sides: \[ x\frac{dv}{dx} = \frac{1}{v} + 1 = \frac{1+v}{v} \] Separate variables: \[ \frac{v}{1+v} dv = \frac{dx}{x} \] \[ \left( 1 - \frac{1}{1+v} \right) dv = \frac{dx}{x} \]
Step 3: Integrate Both Sides \[ \int \left( 1 - \frac{1}{1+v} \right) dv = \int \frac{dx}{x} \] \[ v - \log|1+v| = \log|x| + C \]
Step 4: Back-Substitute \( v = y/x \) \[ \frac{y}{x} - \log\left|1 + \frac{y}{x}\right| = \log|x| + C \] \[ \frac{y}{x} - \log\left|\frac{x+y}{x}\right| = \log|x| + C \] \[ \frac{y}{x} - (\log|x+y| - \log|x|) = \log|x| + C \] \[ \frac{y}{x} - \log|x+y| + \log|x| = \log|x| + C \] \[ \frac{y}{x} = \log|x+y| + C \] Let \( C = \log c \): \[ \frac{y}{x} = \log|x+y| + \log c \] \[ \frac{y}{x} = \log(c|x+y|) \] \[ y = x \log(c|x+y|) \]