What is the solution for y^{''}-4y^'+4y=10e^{2x} ?

The solution for y^{''}-4y^'+4y=10e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+5x^2e^{2x}

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y^{''}-4y^'+4y=10e^{2x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 10 e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 5 x^{2} e^{2 x}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 10 e^{2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 5 x^{2} e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 5 x^{2} e^{2 x}

y^{^{''}} - 2 y^{^{'}} + 3 = e^{t}, y (0) = 1, y^{^{'}} (0) = 2

2 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 = 0

(d^{3} - 3 d^{2} + 4 d - 2) y = x^{3}

y^{^{''}} = 2 y + tan^{3} (x), y (0) = \frac{1}{2} tan (x)

x^{^{''}} + 9 x = 10 cos (2 t)

What is the solution for y^{''}-4y^'+4y=10e^{2x} ?
The solution for y^{''}-4y^'+4y=10e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+5x^2e^{2x}