What is the solution for xy^'=y+x^2sin(x),y(pi)=0 ?

The solution for xy^'=y+x^2sin(x),y(pi)=0 is y=-xcos(x)-x

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xy^'=y+x^2sin(x),y(pi)=0

x y^{^{'}} = y + x^{2} sin (x), y (π) = 0

y = - x cos (x) - x

Solution steps

x y^{'} (x) = y + x^{2} sin (x)

Solve linear ODE:

y = - x cos (x) - x

y = - x cos (x) - x

x \to 0 - lim (x^{2} + 4)

\frac{d}{d x} (e^{- 4 x^{6}})

\frac{d y}{d x} + a y = b

\frac{d}{d x} (ln (\frac{1 - sin ( x )}{1 + sin ( x )}))

n = 0 \sum \infty (\frac{7}{8})^{n}

What is the solution for xy^'=y+x^2sin(x),y(pi)=0 ?
The solution for xy^'=y+x^2sin(x),y(pi)=0 is y=-xcos(x)-x