Question

The degree of the differential equation $x=1+\left(\frac{dy}{dx}\right)+\frac{1}{2!}{\left(\frac{dy}{dx}\right)}^2+\frac{1}{3!}{\left(\frac{dy}{dx}\right)}^3+.........$

• A. 3
• B. 2
• C. 1
• D. not defined

Answer 1

Answer: (C)

$x=1+\left(\frac{dy}{dx}\right)+\frac{1}{2!}{\left(\frac{dy}{dx}\right)}^2+\frac{1}{3!}{\left(\frac{dy}{dx}\right)}^3+\dots$
$\left[\because e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots \right]$
$\therefore x=e^{\frac{dy}{dx}}$
$\Rightarrow {\log }_{e}x=\frac{dy}{dx}$
[After taking log on both sides]
Hence, degree of differential equation is $=1$