What is the solution for y^{''}+y^'-20y=0,y(0)=1,y(0)=22 ?

The solution for y^{''}+y^'-20y=0,y(0)=1,y(0)=22 is y=(1-c_{2})e^{4t}+c_{2}e^{-5t}

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y^{''}+y^'-20y=0,y(0)=1,y(0)=22

y^{^{''}} + y^{^{'}} - 20 y = 0, y (0) = 1, y (0) = 22

y = (1 - c_{2}) e^{4 t} + c_{2} e^{- 5 t}

Solution steps

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

Solve linear ODE:

y = (1 - c_{2}) e^{4 t} + c_{2} e^{- 5 t}

y = (1 - c_{2}) e^{4 t} + c_{2} e^{- 5 t}

20 y^{^{''}} + 60 y = 0

\frac{d ^{2} y}{d x ^{2}} - 36 y = 0

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

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

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (0) = 4

What is the solution for y^{''}+y^'-20y=0,y(0)=1,y(0)=22 ?
The solution for y^{''}+y^'-20y=0,y(0)=1,y(0)=22 is y=(1-c_{2})e^{4t}+c_{2}e^{-5t}