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

The solution for y^'+4y=sin(3t) is y=-(3cos(3t))/(25)+(4sin(3t))/(25)+(c_{1})/(e^{4t)}

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y^'+4y=sin(3t)

y^{^{'}} + 4 y = sin (3 t)

y = - \frac{3 cos ( 3 t )}{25} + \frac{4 sin ( 3 t )}{25} + \frac{c _{1}}{e ^{4 t}}

Solution steps

y^{'} (t) + 4 y = sin (3 t)

Solve linear ODE:

y = - \frac{3 cos ( 3 t )}{25} + \frac{4 sin ( 3 t )}{25} + \frac{c _{1}}{e ^{4 t}}

y = - \frac{3 cos ( 3 t )}{25} + \frac{4 sin ( 3 t )}{25} + \frac{c _{1}}{e ^{4 t}}

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

y = \frac{f ( x )}{x ^{5}}

2 tan^{2} (7 x)

(2 x y + 3) d x + (x^{2} - 1) d y = 0

3 x y^{^{'}} + 7 y = 0

What is the solution for y^'+4y=sin(3t) ?
The solution for y^'+4y=sin(3t) is y=-(3cos(3t))/(25)+(4sin(3t))/(25)+(c_{1})/(e^{4t)}