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

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

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

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

y = e^{t} sin (t)

Solution steps

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

Solve linear ODE:

y = e^{t} sin (t)

y = e^{t} sin (t)

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 13 y = 0

y^{^{''}} - 4 y^{^{'}} - y = 0

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

\frac{d ^{2} x}{d t ^{2}} + 16 x = 0

x^{^{''}} + 10 x^{^{'}} + 25 x = 0

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