y=c^{2}+frac{c}{x} is the solution of the dif | vedclass

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

Question

(B) Given the equation $y=c^{2}+\frac{c}{x}$.Differentiating both sides with respect to $x$,we get:$\frac{d y}{d x} = -\frac{c}{x^{2}}$From this,we ca

Vedclass · Accepted Answer

$x^{4}\left(\frac{d y}{d x}\right)^{2}-x\left(\frac{d y}{d x}\right)-y=0$

Vedclass · Answer

(B) Given the equation $y=c^{2}+\frac{c}{x}$.Differentiating both sides with respect to $x$,we get:$\frac{d y}{d x} = -\frac{c}{x^{2}}$From this,we can express $c$ in terms of $x$ and $\frac{d y}{d x}$:$c = -x^{2} \frac{d y}{d x}$Now,substitute this value of $c$ back into the original equation:$y = (-x^{2} \frac{d y}{d x})^{2} + \frac{-x^{2} \frac{d y}{d x}}{x}$$y = x^{4} (\frac{d y}{d x})^{2} - x \frac{d y}{d x}$Rearranging the terms to form the differential equation:$x^{4} (\frac{d y}{d x})^{2} - x \frac{d y}{d x} - y = 0$Thus,the correct option is $B$.

$y=c^{2}+\frac{c}{x}$ is the solution of the differential equation

Similar Questions

$A$ family of curves whose equation is the general solution of a differential equation having order $1$ and degree $3$ is ($g, a, c$ are arbitrary constants):

If $A$ and $B$ are arbitrary constants,then the differential equation having $y=A e^{-x}+B \cos x$ as its general solution is

The differential equation of the family of curves $y=e^{x}(A \cos x+B \sin x)$,where $A$ and $B$ are arbitrary constants,is

The differential equation of the family of lines having $x$-intercept $a$ and $y$-intercept $b$ is

The differential equation found by the elimination of the arbitrary constant $K$ from the equation $y = (x + K)e^{-x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module