The equation of the curve passing through (1, | vedclass

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

Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The slope of the curve is given by $\frac{dy}{dx} = \frac{y - 1}{x^2 + x}$.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: $\int \frac{dy}{y - 1} = \int (\frac{1}{x} - \frac{1}{x + 1}) dx$.$\ln|y - 1| = \ln|x| - \ln|x + 1| + C$.$\ln|y - 1| = \ln|\frac{x}{x + 1}| + C$,which implies $y - 1 = k \cdot \frac{x}{x + 1}$.Since the curve passes through $(1, 0)$,substitute $x = 1$ and $y = 0$: $0 - 1 = k \cdot \frac{1}{1 + 1} \Rightarrow -1 = \frac{k}{2} \Rightarrow k = -2$.Thus,$y - 1 = -2 \cdot \frac{x}{x + 1} \Rightarrow (y - 1)(x + 1) = -2x$.Therefore,$(y - 1)(x + 1) + 2x = 0$.

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

Similar Questions

The solution of the differential equation $(1+e^{-x})(1+y^2) \frac{dy}{dx} = y^2$ which passes through the point $(0,1)$ is

The general solution of the differential equation $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$ is

If the curve $y = f(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \log_e x) dx, x > 0$, then $f(e)$ is equal to:

The general solution of the differential equation $(x-(x+y) \log (x+y)) dx + x dy = 0$ is

The solution of $x dx + y dy = x^2 y dy - x y^2 dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module