Question ID: #422
Let $y=y(x)$ be the solution of the differential equation $2 \cos x\frac{dy}{dx}=\sin 2x-4y \sin x$, $x\in(0,\frac{\pi}{2})$. If $y(\frac{\pi}{3})=0$, then $y'(\frac{\pi}{4})+y(\frac{\pi}{4})$ is equal to:
Solution:
Dividing the entire equation by $2 \cos x$:
$$ \frac{dy}{dx} = \frac{2 \sin x \cos x}{2 \cos x} – \frac{4y \sin x}{2 \cos x} $$
$$ \frac{dy}{dx} = \sin x – 2y \tan x $$
$$ \frac{dy}{dx} + (2 \tan x)y = \sin x $$
Now, we find the Integrating Factor (I.F.):
$$ I.F. = e^{\int 2 \tan x dx} = e^{2 \ln |\sec x|} = e^{\ln (\sec^2 x)} = \sec^2 x $$
The general solution is given by:
$$ y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + C $$
$$ y \sec^2 x = \int \sin x \sec^2 x dx = \int \sec x \tan x dx $$
$$ y \sec^2 x = \sec x + C $$
$$ y = \cos x + C \cos^2 x $$
Using the initial condition $y(\frac{\pi}{3}) = 0$:
$$ 0 = \cos\frac{\pi}{3} + C \cos^2\frac{\pi}{3} $$
$$ 0 = \frac{1}{2} + C\left(\frac{1}{4}\right) \Rightarrow \frac{C}{4} = -\frac{1}{2} \Rightarrow C = -2 $$
So, the function is $y = \cos x – 2 \cos^2 x$.
We need to find $y'(\frac{\pi}{4}) + y(\frac{\pi}{4})$.
$$ y(\pi/4) = \frac{1}{\sqrt{2}} – 2\left(\frac{1}{2}\right) = \frac{1}{\sqrt{2}} – 1 $$
Differentiating $y$ with respect to $x$:
$$ y’ = -\sin x – 4 \cos x (-\sin x) = -\sin x + 2 \sin 2x $$
$$ y'(\pi/4) = -\frac{1}{\sqrt{2}} + 2(1) = 2 – \frac{1}{\sqrt{2}} $$
Adding the two values:
$$ y'(\pi/4) + y(\pi/4) = \left( 2 – \frac{1}{\sqrt{2}} \right) + \left( \frac{1}{\sqrt{2}} – 1 \right) = 1 $$
Ans. (1)
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