Question

The degree of the differential equation $\sqrt[3]{31+{\left(\frac{dy}{dx}\right)}^{\frac{1}{2}}}=\frac{d^{2}y}{d{x}^2}$ is:

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is D $6$

$\sqrt[3]{31+{\left(\frac{dy}{dx}\right)}^{\frac{1}{2}}}=\frac{d^{2}y}{d{x}^2}$

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

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

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

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

Hence degree (highest power of highest order term) of above differential equation is $6$