What is the solution for y^{''}+2y^'-3y=0,y(0)=1,y^'(0)=3 ?

The solution for y^{''}+2y^'-3y=0,y(0)=1,y^'(0)=3 is y=(1+1/2)e^t-1/2 e^{-3t}

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y^{''}+2y^'-3y=0,y(0)=1,y^'(0)=3

y^{^{''}} + 2 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y = (1 + \frac{1}{2}) e^{t} - \frac{1}{2} e^{- 3 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = (1 + \frac{1}{2}) e^{t} - \frac{1}{2} e^{- 3 t}

y = (1 + \frac{1}{2}) e^{t} - \frac{1}{2} e^{- 3 t}

\frac{d ^{3} y}{d x ^{3}} + 3 \frac{d ^{2} y}{d x ^{2}} - 4 y = 0

y^{^{''}} + k y^{^{'}} + 8 y = 0

100 y^{^{''}} - 140 y^{^{'}} + 85 y = 0, y (2) = 9, y^{^{'}} (2) = 8

x^{^{''}} - 6 x = 0

y^{^{''}} + y^{^{'}} - \frac{10}{3} y = 0

What is the solution for y^{''}+2y^'-3y=0,y(0)=1,y^'(0)=3 ?
The solution for y^{''}+2y^'-3y=0,y(0)=1,y^'(0)=3 is y=(1+1/2)e^t-1/2 e^{-3t}