The solution of the differential equation x f | vedclass

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Question

(A) Given equation: $x \frac{dy}{dx} + y = x^3 y^6$.Divide by $x y^6$: $\frac{1}{y^6} \frac{dy}{dx} + \frac{1}{x y^5} = x^2$.Let $v = y^{-5}$, then $\

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$y^{-5} = \frac{5}{2} x^3 + cx^5$

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(A) Given equation: $x \frac{dy}{dx} + y = x^3 y^6$.Divide by $x y^6$: $\frac{1}{y^6} \frac{dy}{dx} + \frac{1}{x y^5} = x^2$.Let $v = y^{-5}$, then $\frac{dv}{dx} = -5 y^{-6} \frac{dy}{dx}$, so $\frac{1}{y^6} \frac{dy}{dx} = -\frac{1}{5} \frac{dv}{dx}$.Substitute into the equation: $-\frac{1}{5} \frac{dv}{dx} + \frac{v}{x} = x^2$.Multiply by $-5$: $\frac{dv}{dx} - \frac{5}{x} v = -5 x^2$.This is a linear differential equation with integrating factor $IF = e^{\int -\frac{5}{x} dx} = e^{-5 \ln x} = x^{-5}$.Multiply by $IF$: $\frac{d}{dx} (v x^{-5}) = -5 x^2 \cdot x^{-5} = -5 x^{-3}$.Integrate both sides: $v x^{-5} = \int -5 x^{-3} dx = -5 \frac{x^{-2}}{-2} + c = \frac{5}{2} x^{-2} + c$.Multiply by $x^5$: $v = \frac{5}{2} x^3 + c x^5$.Since $v = y^{-5}$, the solution is $y^{-5} = \frac{5}{2} x^3 + c x^5$.

The solution of the differential equation $x \frac{dy}{dx} + y = x^3 y^6$ is:

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