The general solution of the differential equa | vedclass

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(A) Given the differential equation: $(x^3-y^3) dx = (x^2y - xy^2) dy$. Rearranging, we get: $\frac{dy}{dx} = \frac{x^3-y^3}{x^2y - xy^2} = \frac{x^3-

Vedclass · Accepted Answer

$y = x \log(c|x+y|)$

Vedclass · Answer

(A) Given the differential equation: $(x^3-y^3) dx = (x^2y - xy^2) dy$. Rearranging, we get: $\frac{dy}{dx} = \frac{x^3-y^3}{x^2y - xy^2} = \frac{x^3-y^3}{xy(x-y)}$. Since this is a homogeneous equation, let $y = vx$, so $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x^3 - v^3x^3}{x^2(vx) - x(v^2x^2)} = \frac{x^3(1-v^3)}{x^3(v-v^2)} = \frac{1-v^3}{v-v^2} = \frac{(1-v)(1+v+v^2)}{v(1-v)} = \frac{1+v+v^2}{v}$. Then $x \frac{dv}{dx} = \frac{1+v+v^2}{v} - v = \frac{1+v+v^2-v^2}{v} = \frac{1+v}{v}$. Separating variables: $\frac{v}{1+v} dv = \frac{dx}{x}$. Integrating both sides: $\int (1 - \frac{1}{1+v}) dv = \int \frac{dx}{x}$. $v - \log|1+v| = \log|x| + C$. Substituting $v = \frac{y}{x}$: $\frac{y}{x} - \log|1 + \frac{y}{x}| = \log|x| + C$. $\frac{y}{x} - \log|\frac{x+y}{x}| = \log|x| + C$. $\frac{y}{x} - (\log|x+y| - \log|x|) = \log|x| + C$. $\frac{y}{x} = \log|x+y| + C$. Thus, $y = x \log|x+y| + Cx$, which can be written as $y = x \log(c|x+y|)$.

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

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