What is the solution for y^{''}+4y^'+3y=4,e^3ty(0)=2,y^'(0)=7 ?

The solution for y^{''}+4y^'+3y=4,e^3ty(0)=2,y^'(0)=7 is y=c_{1}e^{-t}+c_{2}e^{-3t}+4/3

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y^{''}+4y^'+3y=4,e^3ty(0)=2,y^'(0)=7

y^{^{''}} + 4 y^{^{'}} + 3 y = 4, e^{3} t y (0) = 2, y^{^{'}} (0) = 7

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{4}{3}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 4

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{4}{3}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{4}{3}

y^{^{''}} - 5 y^{^{'}} - 6 y = 6 x^{2} - 4 x - 3

(d^{2} + 4) y = sin^{2} (x)

(d^{2} - 2 d + 2) y = e^{x} tan (x)

y^{^{''}} - y^{^{'}} - 6 y = 11

(d + 2) (d - 1)^{3} y = e^{x}

What is the solution for y^{''}+4y^'+3y=4,e^3ty(0)=2,y^'(0)=7 ?
The solution for y^{''}+4y^'+3y=4,e^3ty(0)=2,y^'(0)=7 is y=c_{1}e^{-t}+c_{2}e^{-3t}+4/3