What is the solution for y^{''}-y=(1-e^2x) ?

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

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y^{''}-y=(1-e^2x)

y^{^{''}} - y = (1 - e^{2} x)

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

Solution steps

y^{''} (x) - y = (1 - e^{2} x)

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = 2 sin (4 x)

y^{^{''}} + 5 y = 6 t^{2} - 9, y (0) = 0, y^{^{'}} (0) = - 3

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

y^{^{''}} + y = cos (0.9 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 5 x e^{2 x} + 10 sin (2 x)

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