Q.20. Let be the solution curve of the differential equation , . If the curve passes through the p

Q: What is the correct answer for this exam Mathematics Q20?

The correct answer is D. Sol. The given differential equation is $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$. Rearrange to separate variables: $(1 + \sin x) \frac{dy}{dx} = -(y + 1) \cos x$ $\frac{dy}{y+1} = -\frac{\cos x}{1 + \sin x} dx$. Integrate both sides: $\int \frac{dy}{y+1} = -\int \frac{\cos x}{1 + \sin x} dx$. Let $u = 1 + \sin x$, then $du = \cos x dx$. $\ln|y+1| = -\ln|1 + \sin x| + C$. $\ln|y+1| + \ln|1

Q: Q.20. Let be the solution curve of the differential equation , . If the curve passes through the point , then a value of is :

The correct answer is D. Sol. The given differential equation is $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$. Rearrange to separate variables: $(1 + \sin x) \frac{dy}{dx} = -(y + 1) \cos x$ $\frac{dy}{y+1} = -\frac{\cos x}{1 + \sin x} dx$. Integrate both sides: $\int \frac{dy}{y+1} = -\int \frac{\cos x}{1 + \sin x} dx$. Let $u = 1 + \sin x$, then $du = \cos x dx$. $\ln|y+1| = -\ln|1 + \sin x| + C$. $\ln|y+1| + \ln|1

Q: What are the options for this this exam Mathematics question?

A. $\frac{\pi}{6}$ · B. $\frac{\pi}{4}$ · C. $\frac{\pi}{3}$ · D. $\frac{\pi}{2}$

Question

Q.20. Let  be the solution curve of the differential equation , . If the curve  passes through the point , then a value of  is :

FirstInTest · Accepted Answer

Sol. The given differential equation is $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.
Rearrange to separate variables:
$(1 + \sin x) \frac{dy}{dx} = -(y + 1) \cos x$
$\frac{dy}{y+1} = -\frac{\cos x}{1 + \sin x} dx$.
Integrate both sides:
$\int \frac{dy}{y+1} = -\int \frac{\cos x}{1 + \sin x} dx$.
Let $u = 1 + \sin x$, then $du = \cos x dx$.
$\ln|y+1| = -\ln|1 + \sin x| + C$.
$\ln|y+1| + \ln|1 + \sin x| = C$.
$\ln|(y+1)(1 + \sin x)| = C$.
$(y+1)(1 + \sin x) = e^C = A$ (where A is a constant).
Given $y(0) = 0$. Substitute $x=0, y=0$:
$(0+1)(1 + \sin 0) = A$.
$(1)(1+0) = A \Rightarrow A = 1$.
So the particular solution is $(y+1)(1 + \sin x) = 1$.
We are given that the curve passes through $(\alpha, -\frac{1}{2})$. Substitute $x=\alpha, y=-\frac{1}{2}$:
$(-\frac{1}{2}+1)(1 + \sin \alpha) = 1$.
$(\frac{1}{2})(1 + \sin \alpha) = 1$.
$1 + \sin \alpha = 2$.
$\sin \alpha = 1$.
The general solution for $\sin \alpha = 1$ is $\alpha = 2n\pi + \frac{\pi}{2}$ for integer $n$.
From the options, $\frac{\pi}{2}$ is a possible value (for $n=0$).

Answer

$\frac{\pi}{6}$

Answer

$\frac{\pi}{4}$

Answer

$\frac{\pi}{3}$

Answer

$\frac{\pi}{2}$

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