What is the solution for y^{''}+y^'-5x=0 ?

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

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y^{''}+y^'-5x=0

y^{^{''}} + y^{^{'}} - 5 x = 0

y = c_{1} + c_{2} e^{- x} + \frac{5 x ^{2}}{2} - 5 x

Solution steps

y^{''} (x) + y^{'} (x) - 5 x = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + \frac{5 x ^{2}}{2} - 5 x

y = c_{1} + c_{2} e^{- x} + \frac{5 x ^{2}}{2} - 5 x

\int_{- \infty}^{x} \frac{1}{x ^{2}} d x

t an g e n t y = 2 + 1 x + 10 x e^{x}, at x = 0

y^{^{''}} + y = cos (2 x)

3 y^{^{''}} - 4 y^{^{'}} - 4 y = 0, y (0) = - 7, y^{^{'}} (0) = - 8

\int (e^{x} + x) d x

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