What is the solution for y^{''}-2y^'+2y=4x+5,y(0)=3,y^'(0)=0 ?

The solution for y^{''}-2y^'+2y=4x+5,y(0)=3,y^'(0)=0 is y=e^x(-3/2 cos(x)-1/2 sin(x))+2x+9/2

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y^{''}-2y^'+2y=4x+5,y(0)=3,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} + 2 y = 4 x + 5, y (0) = 3, y^{^{'}} (0) = 0

y = e^{x} (- \frac{3}{2} cos (x) - \frac{1}{2} sin (x)) + 2 x + \frac{9}{2}

Solution steps

y^{''} (x) - 2 y^{'} (x) + 2 y = 4 x + 5

Solve linear ODE:

y = e^{x} (- \frac{3}{2} cos (x) - \frac{1}{2} sin (x)) + 2 x + \frac{9}{2}

y = e^{x} (- \frac{3}{2} cos (x) - \frac{1}{2} sin (x)) + 2 x + \frac{9}{2}

y^{^{''}} + 1 y = (sec (x)) tan (x)

y^{^{''}} - y^{^{'}} - 2 y = 13 x + 9

x^{^{''}} + 7 x^{^{'}} + 6 x = sin (2 t)

y^{^{''}} - 25 y = e^{5 x}

y^{^{''}} + 2 y = cos (8 t)

What is the solution for y^{''}-2y^'+2y=4x+5,y(0)=3,y^'(0)=0 ?
The solution for y^{''}-2y^'+2y=4x+5,y(0)=3,y^'(0)=0 is y=e^x(-3/2 cos(x)-1/2 sin(x))+2x+9/2