What is the solution for y^{''}+4y^'-5y=0,y(0)=-3,y^'(0)=21 ?

The solution for y^{''}+4y^'-5y=0,y(0)=-3,y^'(0)=21 is y=e^t-4e^{-5t}

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y^{''}+4y^'-5y=0,y(0)=-3,y^'(0)=21

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = - 3, y^{^{'}} (0) = 21

y = e^{t} - 4 e^{- 5 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = e^{t} - 4 e^{- 5 t}

y = e^{t} - 4 e^{- 5 t}

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

x^{^{''}} + 2 x^{^{'}} + 10 x = 0, x^{^{'}} (0) = 1, x (0) = 0

x^{^{''}} + 6 x^{^{'}} + 10 x = 0

y^{^{''}} + 4 y^{^{'}} + 7 y = 0, y (0) = 3, y^{^{'}} (0) = 7

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} + 16 y = 0, y (0) = 5

What is the solution for y^{''}+4y^'-5y=0,y(0)=-3,y^'(0)=21 ?
The solution for y^{''}+4y^'-5y=0,y(0)=-3,y^'(0)=21 is y=e^t-4e^{-5t}