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

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

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

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

x = e^{t} sin (t)

Solution steps

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

Solve linear ODE:

x = e^{t} sin (t)

x = e^{t} sin (t)

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

2 y^{^{''}} - 4 y^{^{'}} + 13 y = 0

20 y^{^{''}} + 3 y^{^{'}} - 2 y = 0

3 \frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 8 y = 0

x^{^{''}} - 6 x^{^{'}} + 10 x = 0

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