What is the solution for y^{''}+8y^'+25y=0,y^'(0)=2 ?

The solution for y^{''}+8y^'+25y=0,y^'(0)=2 is y=e^{-4t}(-(cos(3t)(2-3c_{2}))/4+c_{2}sin(3t))

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y^{''}+8y^'+25y=0,y^'(0)=2

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y^{^{'}} (0) = 2

y = e^{- 4 t} (- \frac{cos ( 3 t ) ( 2 - 3 c _{2} )}{4} + c_{2} sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{- 4 t} (- \frac{cos ( 3 t ) ( 2 - 3 c _{2} )}{4} + c_{2} sin (3 t))

y = e^{- 4 t} (- \frac{cos ( 3 t ) ( 2 - 3 c _{2} )}{4} + c_{2} sin (3 t))

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 5 y = 0

y^{^{'''}} + 3 y^{^{''}} - 18 y^{^{'}} - 40 y = 0

y^{^{''}} + 2 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 3

2 \frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 105 y = 0

6 y^{^{''''}} + 77 y^{^{''}} + 255 y = 0

What is the solution for y^{''}+8y^'+25y=0,y^'(0)=2 ?
The solution for y^{''}+8y^'+25y=0,y^'(0)=2 is y=e^{-4t}(-(cos(3t)(2-3c_{2}))/4+c_{2}sin(3t))