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: Understanding the Concept
The given differential equation is a homogeneous equation because each term in the expressions for $dx$ and $dy$ is of the same degree (degree 3). Homogeneous equations can be solved by making the substitution $y=vx$.
Step 2: Key Formula or Approach
1. Rewrite the equation in the form $\frac{dy}{dx} = f(x,y)$. 2. Substitute $y=vx$, which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$. 3. The equation will transform into a separable variable form in terms of $v$ and $x$. 4. Integrate both sides and substitute back $v=y/x$ to get the general solution.
Step 3: Detailed Explanation
The equation is $(x^3-y^3)dx = (x^2y-xy^2)dy$. \[ \frac{dy}{dx} = \frac{x^3-y^3}{x^2y-xy^2} \] This is a homogeneous equation of degree 3. Let $y=vx$. Then $\frac{dy}{dx} = v+x\frac{dv}{dx}$. \[ v+x\frac{dv}{dx} = \frac{x^3-(vx)^3}{x^2(vx)-x(vx)^2} = \frac{x^3(1-v^3)}{x^3(v-v^2)} = \frac{1-v^3}{v-v^2} \] \[ x\frac{dv}{dx} = \frac{1-v^3}{v-v^2} - v = \frac{1-v^3 - v(v-v^2)}{v-v^2} = \frac{1-v^3-v^2+v^3}{v(1-v)} = \frac{1-v^2}{v(1-v)} \] \[ x\frac{dv}{dx} = \frac{(1-v)(1+v)}{v(1-v)} = \frac{1+v}{v} \] This is a separable equation. \[ \frac{v}{1+v} dv = \frac{1}{x} dx \] To integrate the left side, we rewrite the fraction: \[ \frac{v}{1+v} = \frac{1+v-1}{1+v} = 1 - \frac{1}{1+v} \] Now, integrate both sides: \[ \int \left(1 - \frac{1}{1+v}\right) dv = \int \frac{1}{x} dx \] \[ v - \ln|1+v| = \ln|x| + \ln|c_1| \] \[ v = \ln|x| + \ln|1+v| + \ln|c_1| = \ln|c_1 x(1+v)| \] Now substitute back $v=y/x$: \[ \frac{y}{x} = \ln\left|c_1 x\left(1+\frac{y}{x}\right)\right| = \ln\left|c_1 x\left(\frac{x+y}{x}\right)\right| = \ln|c_1(x+y)| \] Let $c=c_1$. \[ \frac{y}{x} = \ln|c(x+y)| \] This is equivalent to $y = x \ln|c(x+y)|$. In option A, the form is $y=x\log(c|x+y|)$. Using the property $\log(ab)=\log a + \log b$, we see that the constant $c$ can be absorbed differently, so the form is correct. Step 4: Final Answer
The general solution of the differential equation is $y = x\log(c|x+y|)$.