What is the solution for y^{''}-2y^'+2y=2e^t,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-2y^'+2y=2e^t,y(0)=0,y^'(0)=1 is y=e^t(-2cos(t)+sin(t))+2e^t

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y^{''}-2y^'+2y=2e^t,y(0)=0,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

y = e^{t} (- 2 cos (t) + sin (t)) + 2 e^{t}

Solution steps

y^{''} (t) - 2 y^{'} (t) + 2 y = 2 e^{t}

Solve linear ODE:

y = e^{t} (- 2 cos (t) + sin (t)) + 2 e^{t}

y = e^{t} (- 2 cos (t) + sin (t)) + 2 e^{t}

What is the solution for y^{''}-2y^'+2y=2e^t,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-2y^'+2y=2e^t,y(0)=0,y^'(0)=1 is y=e^t(-2cos(t)+sin(t))+2e^t