What is a solution to the differential equation #dy/dx = 20ycos(-5x)#?

Question

1714 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-10T20:03:34

# y = C e^{4 sin 5 x)#

it's separable, so we separate

#dy/dx = 20ycos(-5x)#

#1/y dy/dx = 20cos(-5x)#

we integrate

#int \ 1/y dy/dx \ dx = int \20cos(-5x) \ dx#

or

#int \ 1/y dy =20 int \cos(-5x) \ dx#

#int \ 1/y dy =20 int \cos5x \ dx# #qquad qquad # as #cos (-varphi) = cos varphi#

#ln y = 20 1/5 sin 5 x + C#

#ln y = 4 sin 5 x + C#

# y = e^{4 sin 5 x + C} = C e^{4 sin 5 x)#