What is the solution for y^{''}-4y^'+4y=4x+3 ?

The solution for y^{''}-4y^'+4y=4x+3 is y=c_{1}e^{2x}+c_{2}xe^{2x}+x+7/4

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y^{''}-4y^'+4y=4x+3

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 x + 3

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + x + \frac{7}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + x + \frac{7}{4}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + x + \frac{7}{4}

(d^{2} - 1) y = x sin (x) + (1 + x^{2}) e^{x}

y^{^{''}} + 4 y = sec^{3} (2 x)

3 x^{^{''}} - 30 x^{^{'}} + 75 x = 2 t + 1

y^{^{''}} - 6 y^{^{'}} + 5 y = 5 x^{2} + 3 x - 16 - 9 e^{2 x}

y^{^{''}} - 9 y = (x^{2} - 6) sin (3 x)

What is the solution for y^{''}-4y^'+4y=4x+3 ?
The solution for y^{''}-4y^'+4y=4x+3 is y=c_{1}e^{2x}+c_{2}xe^{2x}+x+7/4