What is the solution for y^'-3y=2sin(3x)-cos(3x) ?

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

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y^'-3y=2sin(3x)-cos(3x)

y^{^{'}} - 3 y = 2 sin (3 x) - cos (3 x)

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

Solution steps

y^{'} (x) - 3 y = 2 sin (3 x) - cos (3 x)

Solve linear ODE:

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

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

\int \frac{6 x + 7}{x ^{2} + 4 x + 29} d x

\frac{d}{d x} (\frac{sin ^{2} ( x )}{cos ( x )})

(- x + 2), (- x^{2} - 2 x), - 2, - 1

(1 + x)^{- \frac{2}{3}}

y = (1 + 2 x)^{10}, (0, 1)

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