What is the solution for y^{''}-2y^'+5y=0,y(0)=0,y^'(0)=2 ?

The solution for y^{''}-2y^'+5y=0,y(0)=0,y^'(0)=2 is y=e^tsin(2t)

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

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

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 6 y = 0

y^{^{'''}} - 9 y^{^{''}} + 28 y^{^{'}} - 20 y = 0

2 y^{^{''}} + 2 y^{^{'}} + 7 y = 0

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

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

What is the solution for y^{''}-2y^'+5y=0,y(0)=0,y^'(0)=2 ?
The solution for y^{''}-2y^'+5y=0,y(0)=0,y^'(0)=2 is y=e^tsin(2t)