Answer to Question #247664 in Differential Equations for JaytheCreator

Question

Solve the differential equation 𝑑𝑦 𝑑𝑥 + 2𝑦 tan 𝑥 = sin 𝑥 , 𝑦 ( 𝜋 3 ) = 0.

Answer 1

Let us solve the differential equation $\frac{dy}{dx}+2y\tan x=\sin x,\ y(\frac{\pi}3)=0.$ Let us divide both parts by $\cos^2 x.$ Then we get the differential equation $\frac{1}{\cos^2 x}y'+\frac{2\sin x}{\cos^3 x}y=\frac{\sin x}{\cos^2 x},$ which is equivalent to $(\frac{1}{\cos^2 x}y)'=\frac{\sin x}{\cos^2 x}.$ It follows that

$\frac{1}{\cos^2 x}y=\int\frac{\sin x}{\cos^2 x}dx=-\int\frac{d(\cos x)}{\cos^2 x}=\frac{1}{\cos x}+C,$ and hence

$y=\cos x+C\cos^2 x$ is the general solution of the differential equation.

Since $y(\frac{\pi}3)=0,$ we get that $0=\cos\frac{\pi}3+C\cos^2\frac{\pi}3=\frac{1}2+C\frac{1}4.$ Consequently, $C=-2.$ We conclude that the solution of $\frac{dy}{dx}+2y\tan x=\sin x,\ y(\frac{\pi}3)=0,$ is

$y=\cos x-2\cos^2 x.$