What is the solution for y^'+y/t =3cos(2t) ?

The solution for y^'+y/t =3cos(2t) is y=(3(2tsin(2t)+cos(2t))}{4t}+\frac{c_{1})/t

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y^'+y/t =3cos(2t)

y^{^{'}} + \frac{y}{t} = 3 cos (2 t)

y = \frac{3 ( 2 t sin ( 2 t ) + cos ( 2 t ) )}{4 t} + \frac{c _{1}}{t}

Solution steps

y^{'} (t) + \frac{y}{t} = 3 cos (2 t)

Solve linear ODE:

y = \frac{3 ( 2 t sin ( 2 t ) + cos ( 2 t ) )}{4 t} + \frac{c _{1}}{t}

y = \frac{3 ( 2 t sin ( 2 t ) + cos ( 2 t ) )}{4 t} + \frac{c _{1}}{t}

\frac{d y}{d x} = \frac{y + x}{y - x}

\frac{d x}{d t} = 5 x

\frac{d}{d x} (b sin (x))

x \to 4 lim (- 2 x^{2})

x \to - \infty lim (- 2 x^{5})

What is the solution for y^'+y/t =3cos(2t) ?
The solution for y^'+y/t =3cos(2t) is y=(3(2tsin(2t)+cos(2t))}{4t}+\frac{c_{1})/t