Question

If \(\frac{d}{dx}(2\frac{d^2y}{dx^2})^3= 7\), then the sum of order and degree of the differential equation is:

• A. 3
• B. 2
• C. 5
• D. 4

Answer 1

The Correct Option is D

The given equation is \(\frac{d}{dx}\left(2\frac{d^2y}{dx^2}\right)^3=7\).

First, let us simplify the expression inside the derivative: \(2\frac{d^2y}{dx^2}=z\). Therefore, \(\left(2\frac{d^2y}{dx^2}\right)=z^{\frac{1}{3}}\) implies \(\left(2\frac{d^2y}{dx^2}\right)^3=z\).

Now, differentiate \(z\) with respect to \(x\):

\(\frac{dz}{dx}=0\).

This simplifies the equation to:

\(\frac{d^3y}{dx^3}=0\).

Here, the order of the differential equation is the highest derivative, which is 3, and the degree is 1 (since \(\frac{d^3y}{dx^3}\) is to the first power).

The sum of the order and the degree is:

\(3+1=4\).

Thus, the sum of the order and degree of the differential equation is 4.