The solution of frac{dy}{dx} = frac{y^2}{xy - | vedclass

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Question

(B) Given the homogeneous differential equation: $\frac{dy}{dx} = \frac{y^2}{xy - x^2}$.Divide numerator and denominator by $x^2$: $\frac{dy}{dx} = \f

Vedclass · Accepted Answer

$e^{y/x} = ky$

Vedclass · Answer

(B) Given the homogeneous differential equation: $\frac{dy}{dx} = \frac{y^2}{xy - x^2}$.Divide numerator and denominator by $x^2$: $\frac{dy}{dx} = \frac{(y/x)^2}{(y/x) - 1}$.Let $v = y/x$,then $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{v^2}{v - 1}$.$x\frac{dv}{dx} = \frac{v^2}{v - 1} - v = \frac{v^2 - v^2 + v}{v - 1} = \frac{v}{v - 1}$.Separating variables: $\frac{v - 1}{v} dv = \frac{dx}{x}$.$(1 - \frac{1}{v}) dv = \frac{dx}{x}$.Integrating both sides: $\int (1 - \frac{1}{v}) dv = \int \frac{dx}{x}$.$v - \ln|v| = \ln|x| + C$.$v = \ln|x| + \ln|v| + C = \ln|xv| + C = \ln|y| + C$.Since $v = y/x$,we have $y/x = \ln|y| + C$.Taking exponential on both sides: $e^{y/x} = e^{\ln|y| + C} = e^C \cdot y$.Let $e^C = k$,then $e^{y/x} = ky$.

Column-$I$	Column-$II$
$A$. Tapetum	$i$. Degenerate during development
$B$. Intine	$ii$. Resistant organic materials
$C$. Vegetative cell	$iii$. Pollen tube
$D$. Sporopollenin	$iv$. Provide nourishment to pollen grains

The solution of $\frac{dy}{dx} = \frac{y^2}{xy - x^2}$ is

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