What is the solution for y^'+y/t =3cos(4t) ?

The solution for y^'+y/t =3cos(4t) is y=(3(4tsin(4t)+cos(4t))}{16t}+\frac{c_{1})/t

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y^'+y/t =3cos(4t)

y^{^{'}} + \frac{y}{t} = 3 cos (4 t)

y = \frac{3 ( 4 t sin ( 4 t ) + cos ( 4 t ) )}{16 t} + \frac{c _{1}}{t}

Solution steps

y^{'} (t) + \frac{y}{t} = 3 cos (4 t)

Solve linear ODE:

y = \frac{3 ( 4 t sin ( 4 t ) + cos ( 4 t ) )}{16 t} + \frac{c _{1}}{t}

y = \frac{3 ( 4 t sin ( 4 t ) + cos ( 4 t ) )}{16 t} + \frac{c _{1}}{t}

y^{^{'}} + y + y^{2} = 0

\frac{d}{d x} (ln (1 + 4 x))

(- 10, - 8), (- 8, - 16)

\frac{d}{d x} (3 e^{5 x})

x = y^{2}, y = x - 2

What is the solution for y^'+y/t =3cos(4t) ?
The solution for y^'+y/t =3cos(4t) is y=(3(4tsin(4t)+cos(4t))}{16t}+\frac{c_{1})/t