What is the solution for y^{''}-2y^'+2y=2x-2,y(pi)=pi,y(0)=0 ?

The solution for y^{''}-2y^'+2y=2x-2,y(pi)=pi,y(0)=0 is y=c_{2}e^xsin(x)+x

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y^{''}-2y^'+2y=2x-2,y(pi)=pi,y(0)=0

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

y = c_{2} e^{x} sin (x) + x

Solution steps

y^{''} (x) - 2 y^{'} (x) + 2 y = 2 x - 2

Solve linear ODE:

y = c_{2} e^{x} sin (x) + x

y = c_{2} e^{x} sin (x) + x

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

y^{^{''}} + y = - 16 cos (3 x)

4 y^{^{''}} - y = e^{\frac{x}{2}} + 7

y^{^{''}} + 6 y^{^{'}} + 13 y = e^{- 3 x} cos (2 x)

y^{^{''}} + y^{^{'}} - 6 y = 2 x^{2} + 1

What is the solution for y^{''}-2y^'+2y=2x-2,y(pi)=pi,y(0)=0 ?
The solution for y^{''}-2y^'+2y=2x-2,y(pi)=pi,y(0)=0 is y=c_{2}e^xsin(x)+x