Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

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

square both sides

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

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

Again square both sides

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

${\left(\frac{d^{2}y}{d{x}^2}\right)}^4-2{\left(\frac{d^{2}y}{d{x}^2}\right)}^2+1-\frac{dy}{dx}=0$

The degree of a differential equation is the highest power of the highest order term. Hence, degree of the given equation is $4$