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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = sin^{22} (x)

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

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 1 + x e^{x}

6 y^{^{''}} + 4 y^{^{'}} - y = 25

10000 x^{^{''}} + 1000 x^{^{'}} + 22 x - 60000 = 0

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