Linear Differential Equations

Linear Differential Equations

Hey JEE aspirants! Differential equations are a crucial part of your JEE Main syllabus, often carrying significant weightage. Among these, linear differential equations are particularly important. They appear frequently and have a straightforward solution method, making them high-scoring if you understand the core concepts. Let's dive in!

What are Linear Differential Equations?

Imagine a situation where the rate of change of a quantity (represented by a function $y$) depends both on the quantity itself and some external factor (represented by a function $x$). Linear differential equations model such scenarios when this dependency is linear. This "linear" nature simplifies the solving process significantly compared to non-linear differential equations.

A linear differential equation is one where the dependent variable ($y$) and its derivatives appear only to the first power and are not multiplied together. Think of it as a "well-behaved" equation where things don't get too complicated. They have several real-world applications, including modeling the growth of populations, the decay of radioactive substances, and the flow of current in electrical circuits.

1. Standard Form

The standard form of a linear differential equation is crucial because it allows us to identify the functions involved and apply the appropriate solution technique. It's like putting everything in the correct format before starting a calculation.

The standard form is:

Here,

• $\frac{dy}{dx}$ represents the first derivative of $y$ with respect to $x$.
• $P(x)$ is a function of $x$ only (it can be a constant). It's the coefficient of the $y$ term.
• $y$ is the dependent variable (the function we're trying to find).
• $Q(x)$ is a function of $x$ only (it can also be a constant). It's the term on the right-hand side, independent of $y$.

Example: Consider the equation $\frac{dy}{dx} + 2xy = x^2$. This is a linear differential equation in standard form, where $P(x) = 2x$ and $Q(x) = x^2$.

Non-Example: The equation $\frac{dy}{dx} + y^2 = x$ is NOT a linear differential equation because $y$ is squared. Similarly, $\frac{dy}{dx} + y\frac{dy}{dx} = x$ is not linear because $y$ is multiplied by its derivative.

4. Recognizing Linear Form

Before applying any solution method, make sure the given equation can be expressed in the standard linear form. Sometimes, the equation might be disguised! This requires some algebraic manipulation.

Tips:

• Check if the dependent variable ($y$) and its derivative ($\frac{dy}{dx}$) appear only to the first power.
• Ensure that $y$ and $\frac{dy}{dx}$ are not multiplied together.
• Verify that the coefficients of $y$ and $\frac{dy}{dx}$ are functions of $x$ only (or constants).

Example: Consider the equation $x\frac{dy}{dx} + y = x^3$. At first glance, it might not seem to be in standard form. However, dividing the entire equation by $x$ (assuming $x \neq 0$) gives us $\frac{dy}{dx} + \frac{1}{x}y = x^2$, which is in standard form, with $P(x) = \frac{1}{x}$ and $Q(x) = x^2$.

Example: The equation $\frac{dy}{dx} = \frac{x^2}{y}$ is not linear. Even though $\frac{dy}{dx}$ appears to the first power, we have $y$ in the denominator on the right side.

2. Integrating Factor

The integrating factor (IF) is the magic ingredient that allows us to solve linear differential equations. It's a function that, when multiplied by the entire equation, makes the left-hand side a perfect derivative. This allows us to integrate both sides easily.

The integrating factor is defined as:

Where $e$ is the base of the natural logarithm, and $\int P(x) \, dx$ is the integral of the function $P(x)$ (which we identified in the standard form).

Explanation: The integrating factor is derived from the product rule of differentiation. The goal is to find a function that, when multiplied by the left side of the standard form equation, will allow it to be written as the derivative of a product.

Example: If $P(x) = 2x$, then the integrating factor is $e^{\int 2x \, dx} = e^{x^2}$.

3. Solution Formula

Once we have the integrating factor, the solution to the linear differential equation is obtained by multiplying the entire equation by the IF and then integrating both sides. Here's the formula:

Where:

• $y$ is the dependent variable.
• IF is the integrating factor we calculated.
• $Q(x)$ is the function on the right-hand side of the standard form.
• $C$ is the constant of integration. Remember to include this!

Derivation:

1. Start with the standard form: $\frac{dy}{dx} + P(x)y = Q(x)$
2. Multiply both sides by the IF: $e^{\int P(x) \, dx} \frac{dy}{dx} + P(x)e^{\int P(x) \, dx}y = Q(x)e^{\int P(x) \, dx}$
3. Notice that the left side is now the derivative of the product $y \cdot e^{\int P(x) \, dx}$: $\frac{d}{dx} \left( y \cdot e^{\int P(x) \, dx} \right) = Q(x)e^{\int P(x) \, dx}$
4. Integrate both sides with respect to $x$: $\int \frac{d}{dx} \left( y \cdot e^{\int P(x) \, dx} \right) dx = \int Q(x)e^{\int P(x) \, dx} dx$
5. This simplifies to: $y \cdot e^{\int P(x) \, dx} = \int Q(x)e^{\int P(x) \, dx} dx + C$
6. Which is our solution formula: $y \times \text{IF} = \int Q(x) \times \text{IF} \, dx + C$

Example: Let's solve $\frac{dy}{dx} + 2xy = x$. We already know $P(x) = 2x$ and $IF = e^{x^2}$. So, using the solution formula:

Let $u = x^2$, then $du = 2x \, dx$, so $\frac{1}{2} du = x\, dx$. Then: $$y \cdot e^{x^2} = \int \frac{1}{2} e^u du + C$$ $$y \cdot e^{x^2} = \frac{1}{2} e^u + C$$ $$y \cdot e^{x^2} = \frac{1}{2} e^{x^2} + C$$

Divide both sides by $e^{x^2}$ to solve for $y$: $$y = \frac{1}{2} + Ce^{-x^2}$$

Linear differential equations are a fundamental tool in mathematics and engineering, and mastering them will not only help you score well in JEE Main but also provide a solid foundation for more advanced topics. Keep practicing, and you'll ace them!