What is the solution for y^{''}+5y^'=0,y^1(0)=1 ?

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

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y^{''}+5y^'=0,y^1(0)=1

y^{^{''}} + 5 y^{^{'}} = 0, y^{1} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + a^{2} \cdot n^{2} \cdot π^{2} \cdot y = 0

3 y^{^{''}} - 4 y^{^{'}} = 0

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 29 y = 0

y^{^{'''}} - 5 y^{^{''}} + 3 y^{^{'}} + y = 0

64 y^{^{''}} + 112 y^{^{'}} + 49 y = 0

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