What is the solution for y^{''}-2y^'-3y=0,y^'(0)=4 ?

The solution for y^{''}-2y^'-3y=0,y^'(0)=4 is y=(e^{3t}(4+c_{2}))/3+c_{2}e^{-t}

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

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

y = \frac{e ^{3 t} ( 4 + c _{2} )}{3} + c_{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{3 t} ( 4 + c _{2} )}{3} + c_{2} e^{- t}

y = \frac{e ^{3 t} ( 4 + c _{2} )}{3} + c_{2} e^{- t}

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

y^{^{''}} - 4 y + 13 y = 0

49 y^{^{'''}} + 56 y^{^{''}} + 16 y^{^{'}} = 0

What is the solution for y^{''}-2y^'-3y=0,y^'(0)=4 ?
The solution for y^{''}-2y^'-3y=0,y^'(0)=4 is y=(e^{3t}(4+c_{2}))/3+c_{2}e^{-t}