Q.Solve the following differential equation: $\frac{dy}{dx} + 2y \tan x = \sin x; \ y=0 \text{ when } x=\frac{\pi}{3}$

First-order linear equation with the initial condition $y(\pi/3)=0$.

Standard form: $\frac{dy}{dx} + 2\tan x\,y = \sin x$, so $P=2\tan x$, $Q=\sin x$.

Integrating factor: $\mu = e^{\int 2\tan x\,dx} = e^{2\log|\sec x|} = \sec^2 x$.

Integrate: $\frac{d}{dx}(y\sec^2 x) = \sin x\sec^2 x = \tan x\sec x$, so …

Spotting the type

$\frac{dy}{dx}+2\tan x\,y=\sin x$ is first-order linear with $P=2\tan x$, $Q=\sin x$.

Integrating factor

so $\mu(x)=e^{\log(\sec^2 x)}=\sec^2 x$.

Multiply and integrate

Since $\frac{d}{dx}\sec x = \sec x\tan x$,

Solve for $y$

Multiply by $\cos^2 x$:

$y = \cos x + C\cos^2 x.$ …

Method: Solve the linear equation, then fix $C$ from the initial value

An initial value problem is solved in two stages: first the general solution by integrating factor, then the constant from the given point.

Steps

Step 1: Standard form and integrating factor.

Write $\dfrac{dy}{dx}+P\,y=Q$ and compute $\text{I.F.}=e^{\int P\,dx}$. For $P=2\tan x$, $\int 2\tan x\,dx=\log(\sec^2x)$, so $\text{I.F.}=\sec^2x$.

Step 2: Multiply and integrate.

Step 3: Simplify to $y=\dots+C(\dots)$. …

Common Mistakes

Mistake 1: Getting the I.F. from $P=2\tan x$ wrong.

Why it's wrong: $\int2\tan x\,dx=-2\log|\cos x|=\log(\sec^2x)$, so I.F. $=\sec^2x$, not $\sec x$ or $e^{2\tan x}$. Correct approach: watch the factor $2$ and the sign.

Mistake 2: Applying the initial condition before finding the general solution.

Why it's wrong: $C$ can only be fixed from $y=\cos x+C\cos^2x$; substituting $y(\pi/3)=0$ too early loses it. Correct approach: get the general solution first, then substitute. …