What is the solution for y^{''}+3y^'+2y=4e^{2x},y(0)=-3,y^'(0)=5 ?

The solution for y^{''}+3y^'+2y=4e^{2x},y(0)=-3,y^'(0)=5 is y=(1-10/3)e^{-x}-e^{-2x}+1/3 e^{2x}

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y^{''}+3y^'+2y=4e^{2x},y(0)=-3,y^'(0)=5

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 e^{2 x}, y (0) = - 3, y^{^{'}} (0) = 5

y = (1 - \frac{10}{3}) e^{- x} - e^{- 2 x} + \frac{1}{3} e^{2 x}

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = 4 e^{2 x}

Solve linear ODE:

y = (1 - \frac{10}{3}) e^{- x} - e^{- 2 x} + \frac{1}{3} e^{2 x}

y = (1 - \frac{10}{3}) e^{- x} - e^{- 2 x} + \frac{1}{3} e^{2 x}

y^{^{'''}} + 4 y^{^{'}} = t e^{- 2 t} + 3 cos (2 t)

y^{^{''}} - y^{^{'}} = 3 e^{t}

y^{^{'''}} - 3 \cdot y^{^{''}} + 2 \cdot y^{^{'}} = - 3 \cdot x^{2} + 6

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

y^{^{'''}} + 6 y^{^{''}} + 12 y^{^{'}} + 8 y = 8 e^{- 2 x}

What is the solution for y^{''}+3y^'+2y=4e^{2x},y(0)=-3,y^'(0)=5 ?
The solution for y^{''}+3y^'+2y=4e^{2x},y(0)=-3,y^'(0)=5 is y=(1-10/3)e^{-x}-e^{-2x}+1/3 e^{2x}