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

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

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y^{''}+y^'-2y=2x+3,y(0)=1,y^'(0)=-4

y^{^{''}} + y^{^{'}} - 2 y = 2 x + 3, y (0) = 1, y^{^{'}} (0) = - 4

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 4 y = 10 t e^{5 t}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 = 0

y^{^{''}} + 2 y^{^{'}} + y = e^{- t} \cdot ln (t)

y^{^{''}} + 9 y = 18 x + 36 sin (x)

y^{^{''}} - y^{^{'}} = e^{- x}

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