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

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

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y^{''}+2y^'-3y=5e^{2x},y(0)=1,y^'(0)=a

y^{^{''}} + 2 y^{^{'}} - 3 y = 5 e^{2 x}, y (0) = 1, y^{^{'}} (0) = a

y = \frac{e ^{x} ( a - 2 )}{4} - \frac{e ^{- 3 x} ( a - 2 )}{4} + e^{2 x}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{x} ( a - 2 )}{4} - \frac{e ^{- 3 x} ( a - 2 )}{4} + e^{2 x}

y = \frac{e ^{x} ( a - 2 )}{4} - \frac{e ^{- 3 x} ( a - 2 )}{4} + e^{2 x}

(D^{2} + 3 D + 2) y = \frac{1}{4 + e ^{x}}

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 6, y (0) = 6, y^{^{'}} (0) = 17

y^{^{''}} - 8 y^{^{'}} + 16 y = (7 t^{4} - 2 t^{3}) e^{4 t}

y^{^{''}} + 4 y + 4 = - 5 e^{2 x}

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