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

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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 $cos (-varphi) = cos varphi$

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

$ln y = 4 sin 5 x + C$

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

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