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

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(A) 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{1}{y^3 x} - \frac{1}{y^4}

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$\sin x + c$

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(A) 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:$\frac{1}{y^3 x} - \frac{1}{y^4} \frac{dy}{dx} = \frac{\cos x}{x^3}$Rearranging the terms:$\frac{1}{y^4} \frac{dy}{dx} - \frac{1}{x y^3} = -\frac{\cos x}{x^3}$Let $t = y^{-3} = \frac{1}{y^3}$. Then,$\frac{dt}{dx} = -3 y^{-4} \frac{dy}{dx}$,which implies $\frac{1}{y^4} \frac{dy}{dx} = -\frac{1}{3} \frac{dt}{dx}$.Substituting this into the equation:$-\frac{1}{3} \frac{dt}{dx} - \frac{t}{x} = -\frac{\cos x}{x^3}$Multiplying by $-3$:$\frac{dt}{dx} + \frac{3}{x} t = \frac{3 \cos x}{x^3}$This is a linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$,where $P(x) = \frac{3}{x}$ and $Q(x) = \frac{3 \cos x}{x^3}$.The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = x^3$.The solution is given by $t \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + c$.$t \cdot x^3 = \int \frac{3 \cos x}{x^3} \cdot x^3 dx + c$$t x^3 = 3 \int \cos x dx + c$$t x^3 = 3 \sin x + c$Since $t = y^{-3}$,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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