What is the solution for y^{''}-4y+9y=0,y^'(0)=1 ?

The solution for y^{''}-4y+9y=0,y^'(0)=1 is y=c_{1}cos(sqrt(5)t)+1/(sqrt(5))sin(sqrt(5)t)

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y^{''}-4y+9y=0,y^'(0)=1

y^{^{''}} - 4 y + 9 y = 0, y^{^{'}} (0) = 1

y = c_{1} cos (5 t) + \frac{1}{5} sin (5 t)

Solution steps

y^{''} (t) - 4 y + 9 y = 0

Solve linear ODE:

y = c_{1} cos (5 t) + \frac{1}{5} sin (5 t)

y = c_{1} cos (5 t) + \frac{1}{5} sin (5 t)

\frac{1}{16} \frac{d ^{2} x}{d t ^{2}} = - 4 x

y^{^{''}} + 16 y = 0, y (\frac{π}{12}) = 2, y^{^{'}} (\frac{π}{12}) = - 4

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

y^{^{''}} + 3 y^{^{''}} + 3 y^{^{'}} + y = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = 3

What is the solution for y^{''}-4y+9y=0,y^'(0)=1 ?
The solution for y^{''}-4y+9y=0,y^'(0)=1 is y=c_{1}cos(sqrt(5)t)+1/(sqrt(5))sin(sqrt(5)t)