What is the solution for y^{'''}+4y^'=3e^x(2x+5) ?

The solution for y^{'''}+4y^'=3e^x(2x+5) is y=c_{1}+c_{2}cos(2x)+c_{3}sin(2x)+(6e^xx)/5+33/25 e^x

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y^{'''}+4y^'=3e^x(2x+5)

y^{^{'''}} + 4 y^{^{'}} = 3 e^{x} (2 x + 5)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{6 e ^{x} x}{5} + \frac{33}{25} e^{x}

Solution steps

y^{'''} (x) + 4 y^{'} (x) = 3 e^{x} (2 x + 5)

Solve linear ODE:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{6 e ^{x} x}{5} + \frac{33}{25} e^{x}

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{6 e ^{x} x}{5} + \frac{33}{25} e^{x}

What is the solution for y^{'''}+4y^'=3e^x(2x+5) ?
The solution for y^{'''}+4y^'=3e^x(2x+5) is y=c_{1}+c_{2}cos(2x)+c_{3}sin(2x)+(6e^xx)/5+33/25 e^x