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

The solution for y^{'''}-4y^'=x+3cos(x) is y=c_{1}+c_{2}e^{-2x}+c_{3}e^{2x}-(x^2)/8-3/5 sin(x)

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

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

y = c_{1} + c_{2} e^{- 2 x} + c_{3} e^{2 x} - \frac{x ^{2}}{8} - \frac{3}{5} sin (x)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + c_{3} e^{2 x} - \frac{x ^{2}}{8} - \frac{3}{5} sin (x)

y = c_{1} + c_{2} e^{- 2 x} + c_{3} e^{2 x} - \frac{x ^{2}}{8} - \frac{3}{5} sin (x)

What is the solution for y^{'''}-4y^'=x+3cos(x) ?
The solution for y^{'''}-4y^'=x+3cos(x) is y=c_{1}+c_{2}e^{-2x}+c_{3}e^{2x}-(x^2)/8-3/5 sin(x)