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Solve the initial value problem.
x y' = y + 7 x^2 sin\(x\), y(4 pi) = 0

User Blunderer
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1 Answer

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xy'=y+7x^2\sin x

xy'-y=7x^2\sin x

\frac1xy'-\frac1{x^2}y=7\sin x

(\mathrm d)/(\mathrm dx)\left[\frac1xy\right]=7\sin x

\frac1xy=\displaystyle\int7\sin x\,\mathrm dx

\frac1xy=-7\cos x+C

y=-7x\cos x+Cx

With the initial value
y(4\pi)=0, we have


0=-28\pi+4\piC\implies C=7

so that the particular solution to the ODE is


y=-7x\cos x+7x
User Danlei
by
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