What is the solution for y^{''}-2y^'=x+2e^x,y(0)=17,y^'(0)= 55/4 ?

The solution for y^{''}-2y^'=x+2e^x,y(0)=17,y^'(0)= 55/4 is y=8e^{2x}+11+(-x^2-x)/4-2e^x

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y^{''}-2y^'=x+2e^x,y(0)=17,y^'(0)= 55/4

y^{^{''}} - 2 y^{^{'}} = x + 2 e^{x}, y (0) = 17, y^{^{'}} (0) = \frac{55}{4}

y = 8 e^{2 x} + 11 + \frac{- x ^{2} - x}{4} - 2 e^{x}

Solution steps

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

Solve linear ODE:

y = 8 e^{2 x} + 11 + \frac{- x ^{2} - x}{4} - 2 e^{x}

y = 8 e^{2 x} + 11 + \frac{- x ^{2} - x}{4} - 2 e^{x}

\frac{1}{s ^{2} - 2 s + 26}

\frac{\partial}{\partial y} (ln (1 + x^{2} + e^{y} + z))

\frac{s ^{2} + 6 s + 2}{s ( s + 1 ) ^{2}}

y^{^{'}} = t y - 2 t + y - 2

3 t^{3} - 2 t^{2} - 7 t + 2

What is the solution for y^{''}-2y^'=x+2e^x,y(0)=17,y^'(0)= 55/4 ?
The solution for y^{''}-2y^'=x+2e^x,y(0)=17,y^'(0)= 55/4 is y=8e^{2x}+11+(-x^2-x)/4-2e^x