First slide

Methods of solving first order first degree differential equations

Question

$If a function y = f (x) satisfies the differential equation$ $f (x) \cdot \sin 2 x - \cos x + (1 + \sin^{2} x) f^{'} (x) = 0 with initial condition y (0) = 0, then the value$ $of f (\frac{π}{6}) is equal to$

Moderate

A

$\frac{1}{5}$
B

$\frac{3}{5}$
C

$\frac{4}{5}$
D

$\frac{2}{5}$

Solution

$\begin{array}{l} The given equation is f (x) \cdot \sin 2 x - \cos x + (1 + \sin^{2} x) f^{'} (x) = 0 \\ \Rightarrow y \sin 2 x - \cos x + (1 + \sin^{2} x) \frac{d y}{d x} = 0 \end{array}$
$\begin{array}{l} \Rightarrow \frac{d y}{d x} + (\frac{\sin 2 x}{1 + \sin^{2} x}) y = \frac{\cos x}{1 + \sin^{2} x} is in the form of \frac{d y}{d x} + P y = Q \\ Where P = \frac{\sin 2 x}{1 + \sin^{2} x} and Q = \frac{\cos x}{1 + \sin^{2} x} \end{array}$
$\begin{array}{l} Integrating Factor (IF) = e^{\int \frac{\sin 2 x}{1 + \sin^{2} x} d x} \\ = e^{\ln (1 + \sin^{2} x)} (∵ \int \frac{f^{'} (x)}{f (x)} d x = \ln | f (x) | + C) \\ = 1 + \sin^{2} x \\ General solution is y \cdot (IF) = \int (IF) Q d x \\ \Rightarrow y (1 + \sin^{2} x) = \int (1 + \sin^{2} x) \frac{\cos x}{(1 + \sin^{2} x)} d x \end{array}$
$\begin{array}{l} \Rightarrow y (1 + \sin^{2} x) = \int \cos x d x = \sin x + C \\ When x = 0, y = 0 \Rightarrow C = 0 \\ When x = \frac{π}{6}, then y (1 + \frac{1}{4}) = \frac{1}{2} \\ \Rightarrow y (\frac{5}{4}) = \frac{1}{2} \\ \Rightarrow y = \frac{2}{5} \\ ∴ f (\frac{π}{6}) = \frac{2}{5} \end{array}$
Therefore, the correct answer is (4).

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