The general solution of the differential equa | vedclass

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Question

(B) Given differential equation: $(y^2 - x^3) dx - xy dy = 0$Divide by $dx$ (assuming $dx 
eq 0$):$y^2 - x^3 - xy \frac{dy}{dx} = 0$$xy \frac{dy}{dx}

Vedclass · Accepted Answer

$y^2 - 2x^3 + cx^2 = 0$

Vedclass · Answer

(B) Given differential equation: $(y^2 - x^3) dx - xy dy = 0$Divide by $dx$ (assuming $dx 
eq 0$):$y^2 - x^3 - xy \frac{dy}{dx} = 0$$xy \frac{dy}{dx} - y^2 = -x^3$Divide by $x$:$y \frac{dy}{dx} - \frac{1}{x} y^2 = -x^2$  ......$(i)$Let $y^2 = v$,then $2y \frac{dy}{dx} = \frac{dv}{dx}$,so $y \frac{dy}{dx} = \frac{1}{2} \frac{dv}{dx}$.Substituting into $(i)$:$\frac{1}{2} \frac{dv}{dx} - \frac{1}{x} v = -x^2$$\frac{dv}{dx} - \frac{2}{x} v = -2x^2$  ......$(ii)$This is a linear differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = -2x^2$.Integrating factor $I.F. = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln |x|} = x^{-2} = \frac{1}{x^2}$.The solution is $v \cdot I.F. = \int Q(x) \cdot I.F. dx + c$.$v \cdot \frac{1}{x^2} = \int (-2x^2) \cdot \frac{1}{x^2} dx + c$$\frac{v}{x^2} = \int -2 dx + c$$\frac{v}{x^2} = -2x + c$Since $v = y^2$,we have $\frac{y^2}{x^2} = -2x + c$.$y^2 = -2x^3 + cx^2$,which implies $y^2 - 2x^3 + cx^2 = 0$.Thus,the correct option is $B$.

The general solution of the differential equation $(y^2 - x^3) dx - xy dy = 0, (x \neq 0)$ is (where $c$ is a constant of integration)

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