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

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

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

linear

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))

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 4, y^{^{'}} (- 1) = 6

\frac{d ^{2} y}{d x ^{2}} = a y

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y^{1} (0) = e^{3} x

5 \frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = 0

2 y^{^{'''}} + 9 y^{^{''}} - 6 y^{^{'}} - 5 y = 0

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