What is the solution for y^'+20y=40sin(60x) ?

The solution for y^'+20y=40sin(60x) is y=(-3cos(60x)+sin(60x))/5+(c_{1})/(e^{20x)}

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y^'+20y=40sin(60x)

y^{^{'}} + 20 y = 40 sin (60 x)

y = \frac{- 3 cos ( 60 x ) + sin ( 60 x )}{5} + \frac{c _{1}}{e ^{20 x}}

Solution steps

y^{'} (x) + 20 y = 40 sin (60 x)

Solve linear ODE:

y = \frac{- 3 cos ( 60 x ) + sin ( 60 x )}{5} + \frac{c _{1}}{e ^{20 x}}

y = \frac{- 3 cos ( 60 x ) + sin ( 60 x )}{5} + \frac{c _{1}}{e ^{20 x}}

\frac{\partial}{\partial x} (sin (8 x) cos (3 y))

y = x^{2}, y = 4 x

\frac{d ^{3}}{d x ^{3}} (x sin (x))

y^{^{'}} = \frac{( y - x )}{x}

f (x) = 6 x^{- 1}

What is the solution for y^'+20y=40sin(60x) ?
The solution for y^'+20y=40sin(60x) is y=(-3cos(60x)+sin(60x))/5+(c_{1})/(e^{20x)}