Number of straight lines which satisfy the di | vedclass

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

Question

(B) Let the equation of the straight line be $y = kx + c$,where $k$ and $c$ are constants. Then,$\frac{dy}{dx} = k$. Substituting this into the given

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the equation of the straight line be $y = kx + c$,where $k$ and $c$ are constants. Then,$\frac{dy}{dx} = k$. Substituting this into the given differential equation $\frac{dy}{dx} + x \left( \frac{dy}{dx} \right)^2 - y = 0$,we get: $k + xk^2 - (kx + c) = 0$ $k + xk^2 - kx - c = 0$ $(k^2 - k)x + (k - c) = 0$ For this to hold for all $x$,the coefficients must be zero: $k^2 - k = 0 \Rightarrow k(k - 1) = 0 \Rightarrow k = 0$ or $k = 1$. If $k = 0$,then $k - c = 0 \Rightarrow c = 0$. The line is $y = 0$. If $k = 1$,then $k - c = 0 \Rightarrow c = 1$. The line is $y = x + 1$. Thus,there are $2$ such straight lines.

Number of straight lines which satisfy the differential equation $\frac{dy}{dx} + x \left( \frac{dy}{dx} \right)^2 - y = 0$ is:

Similar Questions

The number of solutions of $y' = \frac{y + 1}{x - 1}, y(1) = 2$ is

The differential equation $y\frac{dy}{dx} + x = a$ ($a$ is any constant) represents:

The assets of a person are increasing at a rate proportional to the square root of the assets at a given time. His assets increase from Rupees $9$ crores to Rupees $16$ crores in $2$ years,then his assets at the end of $5$ more years will be

The solution of $\frac{d^2 y}{d x^2}=0$ represents

The differential equation $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$ determines a family of circles with

Vedclass Test Series

Exam Paper Generator

Online Exam Module