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)

1

Answer 2

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