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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + y = 7 x sin (x)

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

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

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

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