What is the solution for xy^'=y+4x^2sin(x),y(pi)=0 ?

The solution for xy^'=y+4x^2sin(x),y(pi)=0 is y=-4xcos(x)-4x

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xy^'=y+4x^2sin(x),y(pi)=0

x y^{^{'}} = y + 4 x^{2} sin (x), y (π) = 0

y = - 4 x cos (x) - 4 x

Solution steps

x y^{'} (x) = y + 4 x^{2} sin (x)

Solve linear ODE:

y = - 4 x cos (x) - 4 x

y = - 4 x cos (x) - 4 x

(1 + x^{2}) \frac{d y}{d x} = \frac{x ^{1}}{y}

\frac{d}{d x} (x^{4} - 13 x^{2} + 36)

∣ x ∣, 4 - x^{2}

sin (x), cos (x), \frac{3 π}{4}, \frac{7 π}{6}

\int r^{2} - y^{2} d y

What is the solution for xy^'=y+4x^2sin(x),y(pi)=0 ?
The solution for xy^'=y+4x^2sin(x),y(pi)=0 is y=-4xcos(x)-4x