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

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

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

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

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

\int 3 x^{3} cos (x^{2}) d x

\int \frac{( 5 x + 3 )}{( x ^{3} - 2 x ^{2} - 3 x )} d x

x^{2} + y^{2} = 169, (- 5, 12)

1 - 2 t

x \to \infty lim (π x)

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