What is the solution for y^{''}-4y^'+5y=0,y(0)=2,y^'(0)=1 ?

The solution for y^{''}-4y^'+5y=0,y(0)=2,y^'(0)=1 is y=e^{2t}(2cos(t)-3sin(t))

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y^{''}-4y^'+5y=0,y(0)=2,y^'(0)=1

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y = e^{2 t} (2 cos (t) - 3 sin (t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{2 t} (2 cos (t) - 3 sin (t))

y = e^{2 t} (2 cos (t) - 3 sin (t))

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 16 y = 0

y^{^{'''}} + 4 y^{^{''}} + 5 y^{^{'}} + 2 y = 0

y^{^{''''}} + 20 y^{^{''}} + 100 y = 0

y^{^{''}} - 2 y^{^{'}} + 6 y = 0

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

What is the solution for y^{''}-4y^'+5y=0,y(0)=2,y^'(0)=1 ?
The solution for y^{''}-4y^'+5y=0,y(0)=2,y^'(0)=1 is y=e^{2t}(2cos(t)-3sin(t))