What is the solution for y^{''}-4y^'+29y=0,y(0)=6,y^'(0)=-3 ?

The solution for y^{''}-4y^'+29y=0,y(0)=6,y^'(0)=-3 is y=e^{2t}(6cos(5t)-3sin(5t))

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y^{''}-4y^'+29y=0,y(0)=6,y^'(0)=-3

y^{^{''}} - 4 y^{^{'}} + 29 y = 0, y (0) = 6, y^{^{'}} (0) = - 3

y = e^{2 t} (6 cos (5 t) - 3 sin (5 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{2 t} (6 cos (5 t) - 3 sin (5 t))

y = e^{2 t} (6 cos (5 t) - 3 sin (5 t))

6 y^{^{''}} - 13 y^{^{'}} + 6 y = 0

y^{^{''}} + 5 y^{^{'}} + 8 y = 0

y^{^{'''}} - 19 y^{^{'}} + 30 y = 0

- \frac{d ^{2} y}{d x ^{2}} - \frac{y}{4} = 0

y^{^{''}} + 5 y^{^{'}} - 6 y = 0, y (0) = - 3, y^{^{'}} (0) = 39

What is the solution for y^{''}-4y^'+29y=0,y(0)=6,y^'(0)=-3 ?
The solution for y^{''}-4y^'+29y=0,y(0)=6,y^'(0)=-3 is y=e^{2t}(6cos(5t)-3sin(5t))