What is the solution for y^{''}-2y^'+y=3e^{2x} ?

The solution for y^{''}-2y^'+y=3e^{2x} is y=c_{1}e^x+c_{2}xe^x+3e^{2x}

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + 3 e^{2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + 3 e^{2 x}

y = c_{1} e^{x} + c_{2} x e^{x} + 3 e^{2 x}

y = \frac{4}{x}, (- 5, - \frac{4}{5})

7 x y \frac{d y}{d x} = 1

(- 5, - 7), (4, - 1)

x, at x = 81

e^{r}

What is the solution for y^{''}-2y^'+y=3e^{2x} ?
The solution for y^{''}-2y^'+y=3e^{2x} is y=c_{1}e^x+c_{2}xe^x+3e^{2x}