If x^2 y - x^3 frac{dy}{dx} = y^4 cos x, then | vedclass

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(D) Given the differential equation: $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$.Dividing both sides by $x^3 y^4$, we get:$\frac{x^2 y}{x^3 y^4} - \frac{

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$3 \cos x + c$

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(D) Given the differential equation: $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$.Dividing both sides by $x^3 y^4$, we get:$\frac{x^2 y}{x^3 y^4} - \frac{x^3}{x^3 y^4} \frac{dy}{dx} = \frac{y^4 \cos x}{x^3 y^4}$$\Rightarrow \frac{1}{x y^3} - \frac{1}{y^4} \frac{dy}{dx} = \frac{\cos x}{x^3}$$\Rightarrow -\frac{1}{y^4} \frac{dy}{dx} + \frac{1}{x} y^{-3} = \frac{\cos x}{x^3}$Let $v = y^{-3}$. Then $\frac{dv}{dx} = -3 y^{-4} \frac{dy}{dx}$, which implies $-\frac{1}{y^4} \frac{dy}{dx} = \frac{1}{3} \frac{dv}{dx}$.Substituting this into the equation:$\frac{1}{3} \frac{dv}{dx} + \frac{1}{x} v = \frac{\cos x}{x^3}$Multiplying by $3$: $\frac{dv}{dx} + \frac{3}{x} v = \frac{3 \cos x}{x^3}$.This is a linear differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$, where $P(x) = \frac{3}{x}$ and $Q(x) = \frac{3 \cos x}{x^3}$.Integrating factor ($I$.$F$.) $= e^{\int P(x) dx} = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = x^3$.The solution is $v \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + c$.$v \cdot x^3 = \int \frac{3 \cos x}{x^3} \cdot x^3 dx + c$$v x^3 = 3 \int \cos x dx + c$$v x^3 = 3 \sin x + c$.Since $v = y^{-3}$, we have $x^3 y^{-3} = 3 \sin x + c$.

If $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$, then $x^3 y^{-3}$ is equal to

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