Question

Consider a function y = f(x) satisfying differential equation $\cos xdy = y(\sin x - y)dx, 0 < x < \frac{\pi}{2}$ such that $f(\frac{\pi}{4})=-\sqrt{2}$. Then the value of $\lim_{x \to (\frac{\pi}{2})^{-}} f(x)$ is

• A. 1
• B. -1
• C. 0
• D. $\sqrt{2}$

Answer 1

Answer: (A)

1

Answer 2

The given differential equation is $\cos x dy = y(\sin x - y)dx$. Rearranging, we get $\frac{dy}{dx} = \frac{y(\sin x - y)}{\cos x} = y \tan x - y^2 \sec x$. This is a Bernoulli's differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$, with $P(x) = -\tan x$, $Q(x) = -\sec x$, and $n=2$.

To solve this, we divide by $y^n = y^2$: $y^{-2}\frac{dy}{dx} - y^{-1}\tan x = -\sec x$. Let $v = y^{1-n} = y^{-1}$. Then $\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}$. Substituting these into the equation gives: $-\frac{dv}{dx} - v \tan x = -\sec x$. Multiplying by -1, we get: $\frac{dv}{dx} + v \tan x = \sec x$.

This is a first-order linear differential equation. The integrating factor (I.F.) is: I.F. = $e^{\int P(x) dx} = e^{\int \tan x dx} = e^{\ln|\sec x|} = \sec x$ (since $0 < x < \frac{\pi}{2}$, $\sec x > 0$).

Multiplying the linear equation by the I.F.: $\sec x \frac{dv}{dx} + v \sec x \tan x = \sec^2 x$. The left side is the derivative of the product $v \cdot (\text{I.F.})$: $\frac{d}{dx}(v \sec x) = \sec^2 x$.

Integrating both sides with respect to $x$: $\int \frac{d}{dx}(v \sec x) dx = \int \sec^2 x dx$ $v \sec x = \tan x + C$.

Substitute back $v = \frac{1}{y}$: $\frac{1}{y} \sec x = \tan x + C$. $\frac{1}{y \cos x} = \frac{\sin x}{\cos x} + C = \frac{\sin x + C \cos x}{\cos x}$. $\frac{1}{y} = \sin x + C \cos x$. So, the general solution is $y = f(x) = \frac{1}{\sin x + C \cos x}$.

We are given the initial condition $f(\frac{\pi}{4}) = -\sqrt{2}$. Substitute $x = \frac{\pi}{4}$ and $y = -\sqrt{2}$: $-\sqrt{2} = \frac{1}{\sin(\frac{\pi}{4}) + C \cos(\frac{\pi}{4})}$. $-\sqrt{2} = \frac{1}{\frac{1}{\sqrt{2}} + C \frac{1}{\sqrt{2}}} = \frac{1}{\frac{1}{\sqrt{2}}(1+C)} = \frac{\sqrt{2}}{1+C}$. Dividing both sides by $\sqrt{2}$: $-1 = \frac{1}{1+C}$. $1+C = -1 \implies C = -2$.

Thus, the particular solution is $y = f(x) = \frac{1}{\sin x - 2 \cos x}$.

We need to find the limit of $f(x)$ as $x \to (\frac{\pi}{2})^{-}$. $\lim_{x \to (\frac{\pi}{2})^{-}} f(x) = \lim_{x \to (\frac{\pi}{2})^{-}} \frac{1}{\sin x - 2 \cos x}$. As $x \to (\frac{\pi}{2})^{-}$, $\sin x \to \sin(\frac{\pi}{2}) = 1$ and $\cos x \to \cos(\frac{\pi}{2}) = 0$. The denominator approaches $1 - 2(0) = 1$. Therefore, the limit is $\frac{1}{1} = 1$.