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

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

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

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

y = 2 e^{t} sin (t)

Solution steps

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

Solve linear ODE:

y = 2 e^{t} sin (t)

y = 2 e^{t} sin (t)

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = 8, y^{^{'}} (0) = - 2

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = 3, y^{^{'}} (0) = - 5

\frac{d ^{2} f}{d x ^{2}} + f (x) = 0

6 y^{^{''}} + y^{^{'}} - 2 y = 0, y (1) = 2, y^{^{'}} (1) = - \frac{1}{6}

y^{^{''}} - 2 \cdot y^{^{'}} + y = 0

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