The degree of the differential equation whose | vedclass

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

Question

(A) Given the equation: $y^2 = 8ax + 8a^2$ $(1)$Differentiating with respect to $x$:$2y \frac{dy}{dx} = 8a$$\Rightarrow a = \frac{y}{4} \frac{dy}{dx}$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given the equation: $y^2 = 8ax + 8a^2$ $(1)$Differentiating with respect to $x$:$2y \frac{dy}{dx} = 8a$$\Rightarrow a = \frac{y}{4} \frac{dy}{dx}$Substituting the value of $a$ into equation $(1)$:$y^2 = 8 \left( \frac{y}{4} \frac{dy}{dx} \right) x + 8 \left( \frac{y}{4} \frac{dy}{dx} \right)^2$$y^2 = 2xy \frac{dy}{dx} + 8 \left( \frac{y^2}{16} \right) \left( \frac{dy}{dx} \right)^2$$y^2 = 2xy \frac{dy}{dx} + \frac{y^2}{2} \left( \frac{dy}{dx} \right)^2$Multiplying by $2$ to clear the fraction:$2y^2 = 4xy \frac{dy}{dx} + y^2 \left( \frac{dy}{dx} \right)^2$In this differential equation,the highest order derivative is $\frac{dy}{dx}$,which has an order of $1$. The highest power of the highest order derivative is $2$. Therefore,the degree is $2$.

The degree of the differential equation whose solution is $y^2=8a(x+a)$ is

Similar Questions

Verify that the given function $y=e^{x}+1$ is a solution of the differential equation $y^{\prime \prime}-y^{\prime}=0$.

The differential equation representing the family of ellipses having foci either on the $x$-axis or on the $y$-axis,centered at the origin,and passing through the point $(0,3)$ is:

The elimination of the arbitrary constants $A, B$ and $C$ from $y = A + Bx + C{e^{ - x}}$ leads to the differential equation:

Which of the following differential equations has $y=x$ as one of its particular solutions?

The differential equation whose solution is $y = cx + c - c^3$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module