The solution of the equation x^2 y - x^3 frac | vedclass

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(B) Given the differential equation: $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$. Divide both sides by $x^3 y^4$: $\frac{1}{x^2 y^3} - \frac{1}{y^4} \fra

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$x^3 = 3 y^3 \sin x$

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(B) Given the differential equation: $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$. Divide both sides by $x^3 y^4$: $\frac{1}{x^2 y^3} - \frac{1}{y^4} \frac{dy}{dx} = \frac{\cos x}{x^3}$. Let $v = y^{-3}$,then $\frac{dv}{dx} = -3 y^{-4} \frac{dy}{dx}$,which implies $-\frac{1}{3} \frac{dv}{dx} = y^{-4} \frac{dy}{dx}$. Substituting this into the equation: $\frac{1}{x^3} v + \frac{1}{3} \frac{dv}{dx} = \frac{\cos x}{x^3}$. Multiply by $3$: $\frac{dv}{dx} + \frac{3}{x^3} v = \frac{3 \cos x}{x^3}$. This is a linear differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$. Integrating factor $IF = e^{\int \frac{3}{x^3} dx} = e^{-\frac{3}{2x^2}}$. However,checking the original equation structure,it is a Bernoulli equation. Rearranging $x^3 \frac{dy}{dx} - x^2 y = -y^4 \cos x \implies \frac{dy}{dx} - \frac{1}{x} y = -\frac{y^4 \cos x}{x^3}$. Divide by $y^4$: $y^{-4} \frac{dy}{dx} - \frac{1}{x} y^{-3} = -\frac{\cos x}{x^3}$. Let $v = y^{-3}$,then $\frac{dv}{dx} = -3 y^{-4} \frac{dy}{dx} \implies y^{-4} \frac{dy}{dx} = -\frac{1}{3} \frac{dv}{dx}$. Substituting: $-\frac{1}{3} \frac{dv}{dx} - \frac{1}{x} v = -\frac{\cos x}{x^3} \implies \frac{dv}{dx} + \frac{3}{x} v = \frac{3 \cos x}{x^3}$. $IF = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = x^3$. Solution: $v \cdot x^3 = \int x^3 \cdot \frac{3 \cos x}{x^3} dx = \int 3 \cos x dx = 3 \sin x + C$. $y^{-3} x^3 = 3 \sin x + C$. Given $y(0) = 1$,but the equation is singular at $x=0$. Assuming the form $x^3 = 3 y^3 \sin x$ as a potential solution.

The solution of the equation $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$,where $y(0) = 1$,is

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