Question

If $m$ and $n$ are order and degree of the equation $\left(\frac{d^2y}{dx^2}\right)^3 + 4.\frac{\frac{d^2y}{dx^2}}{\frac{d^3y}{dx^3}} + \frac{d^3y}{dx^3} = x^2 - 1$, then:

• A. $m = 3, n = 3$
• B. $m = 3, n = 2$
• C. $m = 3, n = 5$
• D. $m = 3, n = 1$

Answer 1

Answer: (B)

(B)

Answer 2

To find the order ($m$) and degree ($n$) of the given differential equation, we follow these steps:

The given differential equation is: $\left(\frac{d^2y}{dx^2}\right)^3 + 4.\frac{\frac{d^2y}{dx^2}}{\frac{d^3y}{dx^3}} + \frac{d^3y}{dx^3} = x^2 - 1$

1. Determine the Order (m):

The order of a differential equation is the order of the highest derivative present in the equation. In the given equation, the derivatives are $\frac{d^2y}{dx^2}$ (second order) and $\frac{d^3y}{dx^3}$ (third order). The highest order derivative is $\frac{d^3y}{dx^3}$. Therefore, the order of the equation, $m = 3$.

2. Determine the Degree (n):

The degree of a differential equation is the highest power of the highest order derivative, after the equation has been made free from radicals and fractions as far as derivatives are concerned.

First, we need to clear the fraction involving derivatives. Multiply the entire equation by $\frac{d^3y}{dx^3}$: $\left(\frac{d^2y}{dx^2}\right)^3 \cdot \left(\frac{d^3y}{dx^3}\right) + 4 \cdot \left(\frac{d^2y}{dx^2}\right) + \left(\frac{d^3y}{dx^3}\right)^2 = (x^2 - 1) \cdot \left(\frac{d^3y}{dx^3}\right)$ Rearrange the terms to express it as a polynomial in derivatives: $\left(\frac{d^2y}{dx^2}\right)^3 \cdot \left(\frac{d^3y}{dx^3}\right) + \left(\frac{d^3y}{dx^3}\right)^2 - (x^2 - 1) \cdot \left(\frac{d^3y}{dx^3}\right) + 4 \cdot \left(\frac{d^2y}{dx^2}\right) = 0$ Now, the equation is a polynomial in terms of derivatives. The highest order derivative is $\frac{d^3y}{dx^3}$. We look for the highest power of this highest order derivative.

• In the term $\left(\frac{d^2y}{dx^2}\right)^3 \cdot \left(\frac{d^3y}{dx^3}\right)$, the power of $\frac{d^3y}{dx^3}$ is 1.
• In the term $\left(\frac{d^3y}{dx^3}\right)^2$, the power of $\frac{d^3y}{dx^3}$ is 2.
• In the term $-(x^2 - 1) \cdot \left(\frac{d^3y}{dx^3}\right)$, the power of $\frac{d^3y}{dx^3}$ is 1.

The highest power of the highest order derivative ($\frac{d^3y}{dx^3}$) is 2. Therefore, the degree of the equation, $n = 2$.

So, we have $m = 3$ and $n = 2$.

Comparing this with the given options, the correct option is (B).