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

The solution for y^{''}-y=e^{2x},y(0)=0,y^'(0)=0 is y=(-1/6-1/3)e^x+1/6 e^{-x}+1/3 e^{2x}

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

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

y = (- \frac{1}{6} - \frac{1}{3}) e^{x} + \frac{1}{6} e^{- x} + \frac{1}{3} e^{2 x}

Solution steps

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

Solve linear ODE:

y = (- \frac{1}{6} - \frac{1}{3}) e^{x} + \frac{1}{6} e^{- x} + \frac{1}{3} e^{2 x}

y = (- \frac{1}{6} - \frac{1}{3}) e^{x} + \frac{1}{6} e^{- x} + \frac{1}{3} e^{2 x}

y^{^{''}} - y^{^{'}} + y = e^{2 t}

d (t) = 6000000 \cdot 4^{- 0.15 \cdot 5}

y^{^{''}} - 4 y = e^{2 x} \frac{1}{x}

y^{^{''}} - 3 y^{^{'}} + y = 2 e^{3 t}, y (0) = 0, y^{^{'}} (0) = 4

y^{^{''}} + 3 y^{^{'}} + 2 y = 6 e^{2 t}

What is the solution for y^{''}-y=e^{2x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-y=e^{2x},y(0)=0,y^'(0)=0 is y=(-1/6-1/3)e^x+1/6 e^{-x}+1/3 e^{2x}