What is the solution for y^{''}+8y^'+25y=150,y^'(0)=0,y(0)=0 ?

The solution for y^{''}+8y^'+25y=150,y^'(0)=0,y(0)=0 is y=e^{-4t}(-6cos(3t)-8sin(3t))+6

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y^{''}+8y^'+25y=150,y^'(0)=0,y(0)=0

y^{^{''}} + 8 y^{^{'}} + 25 y = 150, y^{^{'}} (0) = 0, y (0) = 0

y = e^{- 4 t} (- 6 cos (3 t) - 8 sin (3 t)) + 6

Solution steps

y^{''} (t) + 8 y^{'} (t) + 25 y = 150

Solve linear ODE:

y = e^{- 4 t} (- 6 cos (3 t) - 8 sin (3 t)) + 6

y = e^{- 4 t} (- 6 cos (3 t) - 8 sin (3 t)) + 6

y^{^{''}} - 5 y^{^{'}} + 4 y = 2 t e^{t}, y (0) = 1, y^{^{'}} (0) = - 4

\frac{d ^{2} y}{d x ^{2}} + 14 \frac{d y}{d x} + 49 y = 4 e^{3 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e x - 12

y^{^{''}} - y^{^{'}} - 2 y = 26 sin (x) + 40 x cos (x)

y^{^{''}} + 4 y^{^{'}} = x

What is the solution for y^{''}+8y^'+25y=150,y^'(0)=0,y(0)=0 ?
The solution for y^{''}+8y^'+25y=150,y^'(0)=0,y(0)=0 is y=e^{-4t}(-6cos(3t)-8sin(3t))+6