The equation of the curve passing through the | vedclass

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

Question

(A) Given the slope of the curve is $\frac{dy}{dx} = \frac{y - 1}{x^2 + x}$.Separating the variables,we get $\frac{dy}{y - 1} = \frac{dx}{x(x + 1)}$.U

Vedclass · Accepted Answer

$(y - 1)(x + 1) + 2x = 0$

Vedclass · Answer

(A) Given the slope of the curve is $\frac{dy}{dx} = \frac{y - 1}{x^2 + x}$.Separating the variables,we get $\frac{dy}{y - 1} = \frac{dx}{x(x + 1)}$.Using partial fractions,$\frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$.Integrating both sides,$\int \frac{dy}{y - 1} = \int (\frac{1}{x} - \frac{1}{x + 1}) dx$.$\ln|y - 1| = \ln|x| - \ln|x + 1| + C$.$\ln|y - 1| = \ln|\frac{x}{x + 1}| + C$.$y - 1 = k \cdot \frac{x}{x + 1}$,where $k = e^C$.Since the curve passes through $(1, 0)$,substitute $x = 1$ and $y = 0$:$0 - 1 = k \cdot \frac{1}{1 + 1} \implies -1 = \frac{k}{2} \implies k = -2$.Thus,$y - 1 = -2 \cdot \frac{x}{x + 1}$.$(y - 1)(x + 1) = -2x$.$(y - 1)(x + 1) + 2x = 0$.

The equation of the curve passing through the point $(1,0)$ and whose slope is $\frac{y - 1}{x^2 + x}$ is

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $\log _{e}\left(\frac{d y}{d x}\right)=3 x+4 y$,with $y(0)=0$. If $y\left(-\frac{2}{3} \log _{e} 2\right)=\alpha \log _{e} 2$,then the value of $\alpha$ is equal to:

The general solution of $\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y$ is

If $z = z(x)$ and $(2 + \cos x)\frac{dz}{dx} + (\sin x)z = \sin x$,$z(x) > 0$ and $z(\frac{\pi}{2}) = 3$,then $z(\frac{\pi}{3})$ equals -

The solution of the differential equation $\cos(x+y) dy = dx$ given that $y(0) = 0$ is

The particular solution of the differential equation $x \, dy + 2y \, dx = 0$, given that $y = 1$ when $x = 2$, is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module