What is the solution for y^{''}-y^'=1-kx ?

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

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y^{''}-y^'=1-kx

y^{^{''}} - y^{^{'}} = 1 - k x

y = c_{1} e^{x} + c_{2} + \frac{k x ^{2}}{2} + (k - 1) x

Solution steps

y^{''} (x) - y^{'} (x) = 1 - k x

Solve linear ODE:

y = c_{1} e^{x} + c_{2} + \frac{k x ^{2}}{2} + (k - 1) x

y = c_{1} e^{x} + c_{2} + \frac{k x ^{2}}{2} + (k - 1) x

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

y^{^{''}} = 1 - y, y (0) = 0, y (1) = 0

(4 d^{2} + 12 d + 9) y = 144 e^{- 3} x

y^{^{''}} + y^{^{'}} = x^{2} + 2 x + 4

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

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