What is the solution for y^{''}-2y^'+2y=x+1,y(0)=3,y^'(0)=0 ?

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

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y^{''}-2y^'+2y=x+1,y(0)=3,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} + 2 y = x + 1, y (0) = 3, y^{^{'}} (0) = 0

y = e^{x} (2 cos (x) - \frac{5}{2} sin (x)) + \frac{x}{2} + 1

Solution steps

y^{''} (x) - 2 y^{'} (x) + 2 y = x + 1

Solve linear ODE:

y = e^{x} (2 cos (x) - \frac{5}{2} sin (x)) + \frac{x}{2} + 1

y = e^{x} (2 cos (x) - \frac{5}{2} sin (x)) + \frac{x}{2} + 1

d, d (t) = 4000 - 3000 e - 0.06 t

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

y^{^{''}} + y^{^{'}} - 2 y = 2 e^{- 2 x}

\frac{d ^{2} x}{d t ^{2}} + 36 x = 10 sin (4 t), x (0) = 0

y^{^{''}} - 3 y^{^{'}} = 3 e^{3 t}

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