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

The solution for y^{''}+3y^'=0,y(0)=7,y^'(0)=-3 is y=e^{-3t}+6

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

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

y = e^{- 3 t} + 6

Solution steps

y^{''} (t) + 3 y^{'} (t) = 0

Solve linear ODE:

y = e^{- 3 t} + 6

y = e^{- 3 t} + 6

x \frac{d y}{d x} + (3 x + 1) y = e^{- 3 x}

\int \frac{1}{x ^{2} + 2 x + 65} d x

\frac{d}{d x} (3 2 x - 5)

\frac{d}{d x} (\frac{1}{1 + 4 x ^{2}} - arctan (2 x))

e^{- t} sin^{2} (t)

What is the solution for y^{''}+3y^'=0,y(0)=7,y^'(0)=-3 ?
The solution for y^{''}+3y^'=0,y(0)=7,y^'(0)=-3 is y=e^{-3t}+6