What is the solution for y^{''}-2y^'=x+2e^x,y(0)=18,y^'(0)= 31/4 ?

The solution for y^{''}-2y^'=x+2e^x,y(0)=18,y^'(0)= 31/4 is y=5e^{2x}+15+(-x^2-x)/4-2e^x

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y^{''}-2y^'=x+2e^x,y(0)=18,y^'(0)= 31/4

y^{^{''}} - 2 y^{^{'}} = x + 2 e^{x}, y (0) = 18, y^{^{'}} (0) = \frac{31}{4}

y = 5 e^{2 x} + 15 + \frac{- x ^{2} - x}{4} - 2 e^{x}

Solution steps

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

Solve linear ODE:

y = 5 e^{2 x} + 15 + \frac{- x ^{2} - x}{4} - 2 e^{x}

y = 5 e^{2 x} + 15 + \frac{- x ^{2} - x}{4} - 2 e^{x}

\frac{d}{d x} (2 cos (x) + 3 sin (x))

\frac{s + 3}{s ^{2} + 2 s + 2}

5 x^{2} + 8 x - 8

\frac{d}{d u} (u)

\frac{d}{d x} ((6 x - 1) (5 x - 2)^{- 1})

What is the solution for y^{''}-2y^'=x+2e^x,y(0)=18,y^'(0)= 31/4 ?
The solution for y^{''}-2y^'=x+2e^x,y(0)=18,y^'(0)= 31/4 is y=5e^{2x}+15+(-x^2-x)/4-2e^x