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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 9 y = x^{2} + 3 x

- y^{^{''}} + y^{^{'}} + 2 y = 72 t + 400 sin (t) + 36

y^{^{''}} - y^{^{'}} + y = 9 sin (t) - 21 e^{2 t}

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

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (x), y (0) = 1, y^{^{'}} (0) = - 1

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