The equation of the curve which passes throug | vedclass

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

Question

(D) Given the slope of the tangent is $\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2$. This is a homogeneous differential equation. Let

Vedclass · Accepted Answer

$	an ^{-1}\left(\frac{y}{x}ight)=\log |x|$

Vedclass · Answer

(D) Given the slope of the tangent is $\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}ight)^2$. This is a homogeneous differential equation. Let $y = ux$,then $\frac{dy}{dx} = u + x\frac{du}{dx}$. Substituting this into the equation: $u + x\frac{du}{dx} = 1 + u + u^2$. Subtracting $u$ from both sides: $x\frac{du}{dx} = 1 + u^2$. Separating the variables: $\int \frac{du}{1 + u^2} = \int \frac{dx}{x}$. Integrating both sides: $	an^{-1}(u) = \log|x| + C$. Substituting $u = \frac{y}{x}$: $	an^{-1}\left(\frac{y}{x}ight) = \log|x| + C$. Since the curve passes through $(1,0)$,we have $	an^{-1}(0) = \log|1| + C$,which gives $0 = 0 + C$,so $C = 0$. Thus,the equation is $	an^{-1}\left(\frac{y}{x}ight) = \log|x|$.

The equation of the curve which passes through point $(1,0)$ and has a tangent with slope $1+\frac{y}{x}+\left(\frac{y}{x}\right)^{2}$ is

Similar Questions

Find the particular solution satisfying the given condition: $2xy + y^2 - 2x^2 \frac{dy}{dx} = 0$; $y = 2$ when $x = 1$.

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

If $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$ and $y(1) = 1$,then find the value of $x$ that satisfies $y(x) = e$.

One ticket is selected at random from $50$ tickets numbered $\{00, 01, 02, \ldots, 49\}$. Then the probability that the sum of the digits on the selected ticket is $8$,given that the product of these digits is zero,is

The solution of the differential equation $x \frac{dy}{dx} = y - x \tan \left(\frac{y}{x}\right)$ is (Here,$k$ is an arbitrary constant)

Vedclass Test Series

Exam Paper Generator

Online Exam Module