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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''''}} + 8 y^{^{''}} + 16 y = 64 + cos (4 t)

\frac{d ^{2} x}{d t ^{2}} + 25 x = 30 cos (6 t)

y^{^{''}} - 6 y^{^{'}} + 5 y = 3 x + e^{x}, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - y^{^{'}} - 2 y = 2 e^{3 x}

y^{^{''}} + 2 y^{^{'}} + y = 9 - e^{2 x}, y (0) = 0, y^{^{'}} (0) = 1

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