What is the solution for xy^'=y+8x^2sin(x),y(pi)=0 ?

The solution for xy^'=y+8x^2sin(x),y(pi)=0 is y=-8xcos(x)-8x

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xy^'=y+8x^2sin(x),y(pi)=0

x y^{^{'}} = y + 8 x^{2} sin (x), y (π) = 0

y = - 8 x cos (x) - 8 x

Solution steps

x y^{'} (x) = y + 8 x^{2} sin (x)

Solve linear ODE:

y = - 8 x cos (x) - 8 x

y = - 8 x cos (x) - 8 x

3 x^{2} + 4 x y + (2 y + 2 x^{2}) \frac{d y}{d x} = 0

x^{2} \frac{d y}{d x} + 2 x y - y^{3} = 0

\frac{d}{d x} (\frac{10}{x ^{7}})

\frac{d}{d t} (e^{- 5 t})

d u = - e^{- z} d z

What is the solution for xy^'=y+8x^2sin(x),y(pi)=0 ?
The solution for xy^'=y+8x^2sin(x),y(pi)=0 is y=-8xcos(x)-8x