What is the solution for y^{''}-y=e^{2x},y(0)=1 ?

The solution for y^{''}-y=e^{2x},y(0)=1 is y=(-c_{2}+2/3)e^x+c_{2}e^{-x}+1/3 e^{2x}

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

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

y = (- c_{2} + \frac{2}{3}) e^{x} + c_{2} e^{- x} + \frac{1}{3} e^{2 x}

Solution steps

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

Solve linear ODE:

y = (- c_{2} + \frac{2}{3}) e^{x} + c_{2} e^{- x} + \frac{1}{3} e^{2 x}

y = (- c_{2} + \frac{2}{3}) e^{x} + c_{2} e^{- x} + \frac{1}{3} e^{2 x}

y^{^{''}} + y = 3 cos (2 x)

y^{^{''}} + y^{^{'}} = \frac{1}{sin ( t )}

\frac{d ^{2} x}{d t ^{2}} + 5 \frac{d x}{d t} + 3 x = t + 1

y^{^{''}} - 9 y^{^{'}} = x + 1

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

What is the solution for y^{''}-y=e^{2x},y(0)=1 ?
The solution for y^{''}-y=e^{2x},y(0)=1 is y=(-c_{2}+2/3)e^x+c_{2}e^{-x}+1/3 e^{2x}