What is the solution for xy^'=y+5x^2sin(x),y(pi)=0 ?

The solution for xy^'=y+5x^2sin(x),y(pi)=0 is y=-5xcos(x)-5x

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xy^'=y+5x^2sin(x),y(pi)=0

x y^{^{'}} = y + 5 x^{2} sin (x), y (π) = 0

y = - 5 x cos (x) - 5 x

Solution steps

x y^{'} (x) = y + 5 x^{2} sin (x)

Solve linear ODE:

y = - 5 x cos (x) - 5 x

y = - 5 x cos (x) - 5 x

\frac{\partial}{\partial x} (y x^{2})

\int \frac{8}{x + 8 + x} d x

f (z) = e^{\frac{z}{( z - 1 )}}

\frac{d y}{d x} = \frac{x}{2 y ^{2}}

x \to 8 + lim (\frac{7 - x}{8 - x})

What is the solution for xy^'=y+5x^2sin(x),y(pi)=0 ?
The solution for xy^'=y+5x^2sin(x),y(pi)=0 is y=-5xcos(x)-5x