What is the solution for y^{''}-2y^'+5y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-2y^'+5y=0,y(0)=0,y^'(0)=1 is y= 1/2 e^tsin(2t)

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y^{''}-2y^'+5y=0,y(0)=0,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{1}{2} e^{t} sin (2 t)

Solution steps

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

Solve linear ODE:

y = \frac{1}{2} e^{t} sin (2 t)

y = \frac{1}{2} e^{t} sin (2 t)

2 y^{^{''}} + 12 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} - \frac{1}{2} y^{^{'}} + 5 y = 0

x^{^{''}} + 8 x = 0

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

y^{^{'''}} + 3 y^{^{''}} - 10 y^{^{'}} - 24 y = 0

What is the solution for y^{''}-2y^'+5y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-2y^'+5y=0,y(0)=0,y^'(0)=1 is y= 1/2 e^tsin(2t)