The solution of the differential equation (3x | vedclass

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Question

(A) The given differential equation is $(3xy + y^2)dx + (x^2 + xy)dy = 0$. Rearranging the terms,we get $\frac{dy}{dx} = -\frac{3xy + y^2}{x^2 + xy}$.

Vedclass · Accepted Answer

$x^2(2xy + y^2) = c^2$

Vedclass · Answer

(A) The given differential equation is $(3xy + y^2)dx + (x^2 + xy)dy = 0$. Rearranging the terms,we get $\frac{dy}{dx} = -\frac{3xy + y^2}{x^2 + xy}$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these into the equation: $v + x\frac{dv}{dx} = -\frac{3x^2v + x^2v^2}{x^2 + x^2v} = -\frac{3v + v^2}{1 + v}$. $x\frac{dv}{dx} = -\frac{3v + v^2}{1 + v} - v = -\frac{3v + v^2 + v + v^2}{1 + v} = -\frac{2v^2 + 4v}{1 + v} = -\frac{2v(v + 2)}{v + 1}$. Separating the variables: $\frac{v + 1}{v(v + 2)} dv = -\frac{2}{x} dx$. Using partial fractions: $\frac{v + 1}{v(v + 2)} = \frac{A}{v} + \frac{B}{v + 2} \Rightarrow v + 1 = A(v + 2) + Bv$. For $v = 0$,$1 = 2A \Rightarrow A = 1/2$. For $v = -2$,$-1 = -2B \Rightarrow B = 1/2$. So,$\int (\frac{1}{2v} + \frac{1}{2(v + 2)}) dv = -\int \frac{2}{x} dx$. $\frac{1}{2} \ln|v| + \frac{1}{2} \ln|v + 2| = -2 \ln|x| + C'$. $\ln|v(v + 2)| = -4 \ln|x| + 2C' \Rightarrow \ln|v(v + 2)x^4| = C''$. $v(v + 2)x^4 = c^2$. Substituting $v = y/x$: $\frac{y}{x}(\frac{y}{x} + 2)x^4 = c^2 \Rightarrow y(y + 2x)x^2 = c^2 \Rightarrow x^2(y^2 + 2xy) = c^2$.

The solution of the differential equation $(3xy + y^2)dx + (x^2 + xy)dy = 0$ is

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