What is the solution for xy^'=y+6x^2sin(x),y(pi)=0 ?

The solution for xy^'=y+6x^2sin(x),y(pi)=0 is y=-6xcos(x)-6x

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xy^'=y+6x^2sin(x),y(pi)=0

x y^{^{'}} = y + 6 x^{2} sin (x), y (π) = 0

y = - 6 x cos (x) - 6 x

Solution steps

x y^{'} (x) = y + 6 x^{2} sin (x)

Solve linear ODE:

y = - 6 x cos (x) - 6 x

y = - 6 x cos (x) - 6 x

x \to - 1 lim (x^{2} - 3 x - 4)

P = x^{3} - 7 x P, (2, - 6)

x \to 4 lim (\frac{3}{( x - 4 ) ^{2}})

\int cos^{4} (\frac{1}{2} x) d x

\int_{0}^{1} \frac{1}{2} x^{2} d x

What is the solution for xy^'=y+6x^2sin(x),y(pi)=0 ?
The solution for xy^'=y+6x^2sin(x),y(pi)=0 is y=-6xcos(x)-6x