Question

The order and degree of ${\{1+{\left(\frac{dy}{dx}\right)}^2\}}^{\frac{1}{2}}={\left(\frac{d^{2}y}{d{x}^2}\right)}^2$ is ?

• A. $2,2$
• B. $2,4$
• C. $1,2$
• D. $1,4$

Answer 1

The correct option is B $2,4$

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

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

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

We know that, order of a differential equation is the order of the highest order derivative present in the equation and degree of a differential equation is the power of the highest order derivative in the equation.

Here the highest order of the derivative is $2$ and the power of the highest order derivative is $4$.

Thus order and degree is $2,4$.