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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 64 y = 0, y (0) = 0, y^{^{'}} (0) = \frac{64}{5}

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

4 y^{^{''}} + 4 y^{^{'}} + 325 y = 0, y (0) = 3, y^{^{'}} (0) = 2

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

\frac{d ^{2} y}{d x ^{2}} - 25 \cdot y = 0

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