The equation of the curve whose slope is frac | vedclass

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

Question

(B) Given the slope $\frac{dy}{dx} = \frac{y - 1}{x^2 + x} = \frac{y - 1}{x(x + 1)}$.Separating variables: $\int \frac{dy}{y - 1} = \int \frac{dx}{x(x

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the slope $\frac{dy}{dx} = \frac{y - 1}{x^2 + x} = \frac{y - 1}{x(x + 1)}$.Separating variables: $\int \frac{dy}{y - 1} = \int \frac{dx}{x(x + 1)}$.Using partial fractions: $\frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$.Integrating both sides: $\ln|y - 1| = \ln|x| - \ln|x + 1| + C = \ln|\frac{x}{x + 1}| + C$.Since the curve passes through $(1, 0)$, substitute $x = 1, y = 0$: $\ln|0 - 1| = \ln|\frac{1}{1 + 1}| + C \implies 0 = \ln(\frac{1}{2}) + C \implies C = \ln(2)$.Thus, $\ln|y - 1| = \ln|\frac{x}{x + 1}| + \ln(2) = \ln|\frac{2x}{x + 1}|$.Taking exponents: $y - 1 = \frac{2x}{x + 1} \implies (y - 1)(x + 1) = 2x$.

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

Similar Questions

Let $y=y(x)$ be a solution of the differential equation,$\sqrt{1-x^{2}} \frac{dy}{dx}+\sqrt{1-y^{2}}=0, |x| < 1$. If $y\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$,then $y\left(\frac{-1}{\sqrt{2}}\right)$ is equal to

The general solution of the differential equation $\frac{x dy - y dx}{y} = 0$ is . . . . . . .

The general solution of the differential equation $(x-2y+1)dy-(3x-6y+2)dx=0$ is

The equation of a curve passing through the point $(0,1)$,given that the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the $x$-coordinate and the product of $x$ and $y$ coordinates at that point,is

The solution of the equation $\frac{dy}{dx} = \frac{y^2 - y - 2}{x^2 + 2x - 3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module