What is the solution for xy^'=y+9x^2sin(x),y(pi)=0 ?

The solution for xy^'=y+9x^2sin(x),y(pi)=0 is y=-9xcos(x)-9x

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xy^'=y+9x^2sin(x),y(pi)=0

x y^{^{'}} = y + 9 x^{2} sin (x), y (π) = 0

y = - 9 x cos (x) - 9 x

Solution steps

x y^{'} (x) = y + 9 x^{2} sin (x)

Solve linear ODE:

y = - 9 x cos (x) - 9 x

y = - 9 x cos (x) - 9 x

\frac{\partial}{\partial z} (\frac{x}{y} + \frac{y}{z})

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - 8

2 y^{3} d x + 3 y^{2} d y = 0

\int \frac{1}{k x - n x ^{2}} d x

(1.4) (- 5)

What is the solution for xy^'=y+9x^2sin(x),y(pi)=0 ?
The solution for xy^'=y+9x^2sin(x),y(pi)=0 is y=-9xcos(x)-9x