What is the solution for 64y^{''}+96y^'+36y=0,y(0)=1,y^'(0)=8 ?

The solution for 64y^{''}+96y^'+36y=0,y(0)=1,y^'(0)=8 is y=e^{-(3t)/4}+(35e^{-(3t)/4}t)/(4)

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64y^{''}+96y^'+36y=0,y(0)=1,y^'(0)=8

64 y^{^{''}} + 96 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 8

y = e^{- \frac{3 t}{4}} + \frac{35 e ^{- \frac{3 t}{4}} t}{4}

Solution steps

64 y^{''} (t) + 96 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{4}} + \frac{35 e ^{- \frac{3 t}{4}} t}{4}

y = e^{- \frac{3 t}{4}} + \frac{35 e ^{- \frac{3 t}{4}} t}{4}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + 16 y = 0

y^{^{''}} + \frac{32}{13} y^{^{'}} + 32 y = 0

0.2 y^{^{''}} + 14 y^{^{'}} + \frac{1}{0.004} y = 0, y (0) = 0

2 \frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 3 y = 0

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = 1, y^{^{'}} (0) = 25

What is the solution for 64y^{''}+96y^'+36y=0,y(0)=1,y^'(0)=8 ?
The solution for 64y^{''}+96y^'+36y=0,y(0)=1,y^'(0)=8 is y=e^{-(3t)/4}+(35e^{-(3t)/4}t)/(4)