What is the solution for y^{''}+8y^'+25y=0,y(0)=4,y^'(0)=-15 ?

The solution for y^{''}+8y^'+25y=0,y(0)=4,y^'(0)=-15 is y=e^{-4t}(4cos(3t)+1/3 sin(3t))

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y^{''}+8y^'+25y=0,y(0)=4,y^'(0)=-15

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 15

y = e^{- 4 t} (4 cos (3 t) + \frac{1}{3} sin (3 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{- 4 t} (4 cos (3 t) + \frac{1}{3} sin (3 t))

y = e^{- 4 t} (4 cos (3 t) + \frac{1}{3} sin (3 t))

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

y^{^{''}} - 4 y^{^{'}} - a^{2} y = 0

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

y^{^{'''}} - 2 y^{^{''}} - 15 y^{^{'}} = 0

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

What is the solution for y^{''}+8y^'+25y=0,y(0)=4,y^'(0)=-15 ?
The solution for y^{''}+8y^'+25y=0,y(0)=4,y^'(0)=-15 is y=e^{-4t}(4cos(3t)+1/3 sin(3t))