Let y = y(x) be the solution curve of the dif | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given differential equation is $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.Rearranging the terms to separate the variables, we get $\fra

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) The given differential equation is $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.Rearranging the terms to separate the variables, we get $\frac{dy}{y+1} = -\frac{\cos x}{1+\sin x} dx$.Integrating both sides, we have $\int \frac{dy}{y+1} = -\int \frac{\cos x}{1+\sin x} dx$.This yields $\ln|y+1| = -\ln|1+\sin x| + C$, which simplifies to $\ln|y+1| + \ln|1+\sin x| = C$.Using the property of logarithms, we get $(y+1)(1+\sin x) = K$, where $K = e^C$.Given the initial condition $y(0) = 0$, we substitute $x = 0$ and $y = 0$ into the equation: $(0+1)(1+\sin 0) = K$, which gives $1(1+0) = K$, so $K = 1$.Thus, the particular solution is $(y+1)(1+\sin x) = 1$.If the curve passes through $(\alpha, -\frac{1}{2})$, we substitute $x = \alpha$ and $y = -\frac{1}{2}$ into the equation:$(-\frac{1}{2} + 1)(1+\sin \alpha) = 1$.$\frac{1}{2}(1+\sin \alpha) = 1$.$1+\sin \alpha = 2$.$\sin \alpha = 1$.Therefore, $\alpha = \frac{\pi}{2}$.

Let $y = y(x)$ be the solution curve of the differential equation $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$ with the condition $y(0) = 0$. If the curve $y = y(x)$ passes through the point $(\alpha, -\frac{1}{2})$, then a value of $\alpha$ is:

Similar Questions

If $y = f(x)$ is a monotonically increasing function such that $(\frac{dy}{dx})^2 = 6 - \frac{dy}{dx}$ and $y(0) = 5$, then $y(3) = ...$

The general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is

If $x^{3} dy + xy dx = x^{2} dy + 2y dx$,$y(2) = e$ and $x > 1$,then $y(4)$ is equal to

For the differential equation $x y \frac{dy}{dx} = (x+2)(y+2)$,find the solution curve passing through the point $(1, -1)$.

Let $y=y(x)$ be the solution of the differential equation $e^{x} \sqrt{1-y^{2}} dx + \frac{y}{x} dy = 0$,with $y(1) = -1$. Then the value of $(y(3))^{2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module