What is the solution for 4y^{''}+20y^'+25y=0,y(0)=1,y^'(0)=2 ?

The solution for 4y^{''}+20y^'+25y=0,y(0)=1,y^'(0)=2 is y=e^{-(5t)/2}+(9e^{-(5t)/2}t)/(2)

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4y^{''}+20y^'+25y=0,y(0)=1,y^'(0)=2

4 y^{^{''}} + 20 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = e^{- \frac{5 t}{2}} + \frac{9 e ^{- \frac{5 t}{2}} t}{2}

Solution steps

4 y^{''} (t) + 20 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} + \frac{9 e ^{- \frac{5 t}{2}} t}{2}

y = e^{- \frac{5 t}{2}} + \frac{9 e ^{- \frac{5 t}{2}} t}{2}

(D^{3} - 3 D^{2} + 4) y = 0

y^{^{''}} - 24 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = 22

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

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

y^{^{'''}} - y^{^{''}} + 4 y^{^{'}} + 6 y = 0

What is the solution for 4y^{''}+20y^'+25y=0,y(0)=1,y^'(0)=2 ?
The solution for 4y^{''}+20y^'+25y=0,y(0)=1,y^'(0)=2 is y=e^{-(5t)/2}+(9e^{-(5t)/2}t)/(2)