What is the solution for y^{'''}-4y^{''}=2x+3 ?

The solution for y^{'''}-4y^{''}=2x+3 is y=c_{1}e^{4x}+c_{2}+c_{3}x-(x^3)/(12)-(7x^2)/(16)

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

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

y = c_{1} e^{4 x} + c_{2} + c_{3} x - \frac{x ^{3}}{12} - \frac{7 x ^{2}}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} + c_{3} x - \frac{x ^{3}}{12} - \frac{7 x ^{2}}{16}

y = c_{1} e^{4 x} + c_{2} + c_{3} x - \frac{x ^{3}}{12} - \frac{7 x ^{2}}{16}

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3} t

(d^{2} - d - 2) y = 6 e^{\frac{- t}{2}}

d, d (q) = - 0.4 q + 23

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

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

What is the solution for y^{'''}-4y^{''}=2x+3 ?
The solution for y^{'''}-4y^{''}=2x+3 is y=c_{1}e^{4x}+c_{2}+c_{3}x-(x^3)/(12)-(7x^2)/(16)