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: $x^2dy + xydx - 1 = 0$.

Answer 2

Step 1: Understanding Linear Differential Equations
A differential equation is said to be linear if the dependent variable and all its derivatives appear to the power of one (i.e., they are not multiplied together or raised to any power other than one). There should be no products or nonlinear functions (like sin, cos, exponential of the dependent variable) of the dependent variable or its derivatives.

Step 2: Analyze the given equations
To determine which equation is linear, check whether the dependent variable \( y \) and its derivatives appear in a linear form (first power, not multiplied by each other or raised to powers).

Step 3: Examine the equation \( x^2 dy + xy dx - 1 = 0 \)
Rewrite it as:
\( x^2 \frac{dy}{dx} + xy = 1 \)
Here, \(\frac{dy}{dx}\) and \( y \) appear in the first power and are not multiplied together. There are no nonlinear terms involving \( y \) or its derivative.

Step 4: Conclusion
Since the equation can be written in the form:
\( x^2 \frac{dy}{dx} + x y = 1 \),
which is linear in \( y \) and \( \frac{dy}{dx} \), the equation \( x^2 dy + x y dx - 1 = 0 \) is a linear differential equation.