What is the solution for y^{''}+7y^'=0,y(0)=8,y^'(0)=-3 ?

The solution for y^{''}+7y^'=0,y(0)=8,y^'(0)=-3 is y=8-3/7+3/7 e^{-7t}

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y^{''}+7y^'=0,y(0)=8,y^'(0)=-3

y^{^{''}} + 7 y^{^{'}} = 0, y (0) = 8, y^{^{'}} (0) = - 3

y = 8 - \frac{3}{7} + \frac{3}{7} e^{- 7 t}

Solution steps

y^{''} (t) + 7 y^{'} (t) = 0

Solve linear ODE:

y = 8 - \frac{3}{7} + \frac{3}{7} e^{- 7 t}

y = 8 - \frac{3}{7} + \frac{3}{7} e^{- 7 t}

y^{^{''}} + 4 y^{^{'}} + y 32 = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y^{^{'}} (0) = 4

36 \frac{d ^{2} y}{d x ^{2}} + 12 \frac{d y}{d x} + y = 0

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d t ^{2}} - 10 \frac{d y}{d t} = 0

What is the solution for y^{''}+7y^'=0,y(0)=8,y^'(0)=-3 ?
The solution for y^{''}+7y^'=0,y(0)=8,y^'(0)=-3 is y=8-3/7+3/7 e^{-7t}