What is the solution for y^{''}-2y^'+2y=0,y(0)=0,y^'(0)=7 ?

The solution for y^{''}-2y^'+2y=0,y(0)=0,y^'(0)=7 is y=7e^tsin(t)

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y^{''}-2y^'+2y=0,y(0)=0,y^'(0)=7

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 7

y = 7 e^{t} sin (t)

Solution steps

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

Solve linear ODE:

y = 7 e^{t} sin (t)

y = 7 e^{t} sin (t)

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

100 y^{^{''}} + 80 y^{^{'}} + 65 y = 0

y^{^{''}} + y = y^{^{'}}

y^{^{''}} + y = 0, y^{^{'}} (\frac{π}{2}) = 0, y (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = 0, y (0) = 0

What is the solution for y^{''}-2y^'+2y=0,y(0)=0,y^'(0)=7 ?
The solution for y^{''}-2y^'+2y=0,y(0)=0,y^'(0)=7 is y=7e^tsin(t)