What is the solution for y^{''}-2y^'+10y=e^xcos(3x) ?

The solution for y^{''}-2y^'+10y=e^xcos(3x) is y=e^x(c_{1}cos(3x)+c_{2}sin(3x))+(e^xxsin(3x))/6

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y^{''}-2y^'+10y=e^xcos(3x)

y^{^{''}} - 2 y^{^{'}} + 10 y = e^{x} cos (3 x)

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + 10 y = e^{x} cos (3 x)

Solve linear ODE:

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

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

\int sin (u^{2}) d u

\int \frac{37}{x ^{3} - 27} d x

\frac{\partial}{\partial y} (\frac{2 y}{x ^{2} + y ^{2}})

\frac{x ^{2}}{8 + x}

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What is the solution for y^{''}-2y^'+10y=e^xcos(3x) ?
The solution for y^{''}-2y^'+10y=e^xcos(3x) is y=e^x(c_{1}cos(3x)+c_{2}sin(3x))+(e^xxsin(3x))/6