Question

Which one of the following is a linear differential equation?

• A. $\frac{dx}{dy} + y^2 = e^x$
• B. $dr + (2r^2 \cot\theta + \sin 2\theta)d\theta = 0$
• C. $\frac{dy}{dx} = e^{x-y}(e^x - e^{-y})$
• D. $x^2dy + xydx - 1 = 0$

Hint:

Identifying Linear Differential Equations:

• No powers or products of dependent variable.
• Should be first-degree in $y$, $dy/dx$.
• Coefficients depend only on the independent variable.

Answer 1

The Correct Option is D

To identify a linear differential equation among the given options, we must understand that a linear differential equation can be generally expressed in the form:

L(dy/dx) + P(x)y = Q(x)

Where L, P, and Q are functions of x only, and the highest derivative's power is 1. Let's evaluate each option:

1. $\frac{dx}{dy} + y^2 = e^x$
   This is not linear as the term $y^2$ involves a power of y which is not 1.
2. $dr + (2r^2 \cot\theta + \sin 2\theta)d\theta = 0$
   This equation is not linear due to the term $2r^2$, which involves a power of r greater than 1.
3. $\frac{dy}{dx} = e^{x-y}(e^x - e^{-y})$
   This expression is nonlinear as it involves exponential terms like $e^{-y}$ which cannot be expressed as a linear relationship.
4. $x^2dy + xydx - 1 = 0$
   Rewriting in the form $P(x,y)dx + Q(x,y)dy = 0$, we have $x^2dy + xydx = 1$. This rearranges to:
   
   $\frac{dy}{dx} = \frac{-xy}{x^2} + \frac{1}{x^2}$
   
   which simplifies to:
   
   $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x^2}$
   
   This is a linear equation in the form L(dy/dx) + P(x)y = Q(x) where L = 1, P(x) = 1/x, and Q(x) = 1/x².

Therefore, the correct linear differential equation is: $x^2dy + xydx - 1 = 0$.