# Solve the following differential equation : $\large\frac{dy}{dx}$$+2y\tan\: x = \sin\: x. given that $$y=0, \: when\: x = \large\frac{\pi}{3}.$$ ## 1 Answer Toolbox: • To solve the first order linear differential equation of the form \large\frac{dy}{dx}$$ + Py = Q$
• Write the given equation in the form of $\large\frac{dy}{dx}$$+ Py = Q • Find the integrating factor (I.F) = e^{\int Pdx}. • Write the solution as y(I.F) = integration of Q(I.F) dx + C Step 1: Using the information in the tool box, \large\frac{dy}{dx}$$ +2y\tan x = \sin x$
Here $P = 2\tan x$ and $Q = \sin x$
To find the integral factor
$\int P dx = \int 2\tan xdx$
$\qquad\;\;=2 \log|\sec x|$ or $log|\sec^2x|$
Hence $I.F = e^{\large\log|\sec^2x|} = \sec^2x$
Step 2:
Hence the required solution is $y\sec^2x = \int\sin x(\sec^2x)dx + C$ $\int\sin x(\sec^2x)dx = \int\sec x.\tan x dx = \sec x$
Hence the require solution is $y\sec^2x = \sec x + C$
It is given $y = 0$ and $x = \large\frac{\pi}{3}$
substituting these values to evaluate C
$0.\sec^2(\large\frac{\pi}{3}) =$$\sec(\large\frac{\pi}{3})$$ + C$