Question

The degree of the differential equation $\sqrt{1+{\left(\frac{dy}{dx}\right)}^{\frac{1}{3}}}=\frac{d^{2}y}{d{x}^2}$ is:

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is D $6$

$\sqrt{1+{\left(\frac{dy}{dx}\right)}^{1/3}}=\frac{d^{2}y}{d{x}^2}$

$\Rightarrow 1+{\left(\frac{dy}{dx}\right)}^{1/3}={\left(\frac{d^{2}y}{d{x}^2}\right)}^2$, square both sides

$\Rightarrow {\left(\frac{dy}{dx}\right)}^{1/3}={\left(\frac{d^{2}y}{d{x}^2}\right)}^2-1$

$\Rightarrow \left(\frac{dy}{dx}\right)={({\left(\frac{d^{2}y}{d{x}^2}\right)}^2-1)}^3$, cube both sides

$\Rightarrow {\left(\frac{d^{2}y}{d{x}^2}\right)}^6-3{\left(\frac{d^{2}y}{d{x}^2}\right)}^2+3\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}-1=0$

Hence degree is $6$