Question #247664

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


1
Expert's answer
2021-10-11T16:28:59-0400

Let us solve the differential equation dydx+2ytanx=sinx, y(π3)=0.\frac{dy}{dx}+2y\tan x=\sin x,\ y(\frac{\pi}3)=0. Let us divide both parts by cos2x.\cos^2 x. Then we get the differential equation 1cos2xy+2sinxcos3xy=sinxcos2x,\frac{1}{\cos^2 x}y'+\frac{2\sin x}{\cos^3 x}y=\frac{\sin x}{\cos^2 x}, which is equivalent to (1cos2xy)=sinxcos2x.(\frac{1}{\cos^2 x}y)'=\frac{\sin x}{\cos^2 x}. It follows that

1cos2xy=sinxcos2xdx=d(cosx)cos2x=1cosx+C,\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=cosx+Ccos2xy=\cos x+C\cos^2 x is the general solution of the differential equation.

Since y(π3)=0,y(\frac{\pi}3)=0, we get that 0=cosπ3+Ccos2π3=12+C14.0=\cos\frac{\pi}3+C\cos^2\frac{\pi}3=\frac{1}2+C\frac{1}4. Consequently, C=2.C=-2. We conclude that the solution of dydx+2ytanx=sinx, y(π3)=0,\frac{dy}{dx}+2y\tan x=\sin x,\ y(\frac{\pi}3)=0, is

y=cosx2cos2x.y=\cos x-2\cos^2 x.


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