What is the solution for y^{''}-6y^'+10y=e^{3x}cos(x) ?

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

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y^{''}-6y^'+10y=e^{3x}cos(x)

y^{^{''}} - 6 y^{^{'}} + 10 y = e^{3 x} cos (x)

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

Solution steps

y^{''} (x) - 6 y^{'} (x) + 10 y = e^{3 x} cos (x)

Solve linear ODE:

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

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

y^{^{''}} + y = x^{2} + sin (x)

y^{^{''}} - 4 y^{^{'}} - 21 y = 320 e^{- t}

6 y^{^{''}} - y^{^{'}} - y = 50 e^{\frac{x}{2}} + 4 x^{2} - 5

\frac{d ^{2} y}{d x ^{2}} + 5 y - sin (π x) = 0

y^{^{''}} - 10 y^{^{'}} + 9 y = 5 t, y (0) = - 1

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