What is the solution for 4y^{''}-3y^'=4x-6 ?

The solution for 4y^{''}-3y^'=4x-6 is y=c_{1}e^{(3x)/4}+c_{2}-(2x^2)/3+(2x)/9

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4y^{''}-3y^'=4x-6

4 y^{^{''}} - 3 y^{^{'}} = 4 x - 6

y = c_{1} e^{\frac{3 x}{4}} + c_{2} - \frac{2 x ^{2}}{3} + \frac{2 x}{9}

Solution steps

4 y^{''} (x) - 3 y^{'} (x) = 4 x - 6

Solve linear ODE:

y = c_{1} e^{\frac{3 x}{4}} + c_{2} - \frac{2 x ^{2}}{3} + \frac{2 x}{9}

y = c_{1} e^{\frac{3 x}{4}} + c_{2} - \frac{2 x ^{2}}{3} + \frac{2 x}{9}

y^{^{''}} - 9 y^{^{'}} + 8 y = x, y (0) = e^{x}

y^{^{''}} + 2 y^{^{'}} + 4 y = t + cos (2 t)

y^{^{''}} - 4 y^{^{'}} + 4 y = 12 t e^{2 t} (t + 1)

y^{^{''}} - 5 y^{^{'}} + 6 y = 4 e^{4 x}, y (0) = 0, y^{^{'}} (0) = 1

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

What is the solution for 4y^{''}-3y^'=4x-6 ?
The solution for 4y^{''}-3y^'=4x-6 is y=c_{1}e^{(3x)/4}+c_{2}-(2x^2)/3+(2x)/9