What is the solution for 2y^'-y=4asin(3t) ?

The solution for 2y^'-y=4asin(3t) is y=-(4a(6cos(3t)+sin(3t)))/(37)+c_{1}e^{1/2 t}

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2y^'-y=4asin(3t)

2 y^{^{'}} - y = 4 a sin (3 t)

y = - \frac{4 a ( 6 cos ( 3 t ) + sin ( 3 t ) )}{37} + c_{1} e^{\frac{1}{2} t}

Solution steps

2 y^{'} (t) - y = 4 a sin (3 t)

Solve linear ODE:

y = - \frac{4 a ( 6 cos ( 3 t ) + sin ( 3 t ) )}{37} + c_{1} e^{\frac{1}{2} t}

y = - \frac{4 a ( 6 cos ( 3 t ) + sin ( 3 t ) )}{37} + c_{1} e^{\frac{1}{2} t}

G (s) = \frac{8 s ^{2}}{7 e ^{s} + s}

\frac{( x ^{3} - 6 x ) ^{9}}{9}

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

\frac{\partial}{\partial x} (2 x^{3} y + 4)

x \to 3 + lim (\frac{1}{x - 3})

What is the solution for 2y^'-y=4asin(3t) ?
The solution for 2y^'-y=4asin(3t) is y=-(4a(6cos(3t)+sin(3t)))/(37)+c_{1}e^{1/2 t}