Bernoulli's Equation and Applications

Hey JEE aspirants! Differential equations might seem daunting, but they are super useful in modeling real-world phenomena, from population growth to radioactive decay. Bernoulli's equation is a clever extension of linear differential equations that pops up frequently in JEE Main. Mastering it will not only boost your score but also give you a deeper understanding of mathematical modeling.

What is Bernoulli's Equation?

A Bernoulli equation is a type of nonlinear first-order ordinary differential equation that can be transformed into a linear differential equation using a suitable substitution. It has the general form:

\frac{dy}{dx} + P(x)y = Q(x)y^n

Here, $P(x)$ and $Q(x)$ are functions of $x$ , and $n$ is any real number except 0 and 1 (because if $n$ were 0 or 1, the equation would already be linear!).

The Substitution Trick

The magic of Bernoulli's equation lies in a clever substitution that transforms it into a linear form. We use the substitution:

v = y^{1-n}

Let's see how this works. We need to find $\frac{dv}{dx}$ in terms of $\frac{dy}{dx}$ . Differentiating $v$ with respect to $x$ , we get:

\frac{dv}{dx} = (1-n)y^{-n} \frac{dy}{dx}

Now, multiply the original Bernoulli equation by $(1-n)y^{-n}$ :

(1-n)y^{-n} \frac{dy}{dx} + (1-n)P(x)y^{1-n} = (1-n)Q(x)

Notice that the first term is exactly $\frac{dv}{dx}$ , and we defined $v = y^{1-n}$ . So we have:

\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)

This is a linear differential equation in $v$ ! We can solve this using the integrating factor method that you already know.

Converting to Linear Form: An Example

Let's say we have the equation: $\frac{dy}{dx} + \frac{y}{x} = x^2 y^3$ . Here, $P(x) = \frac{1}{x}$ , $Q(x) = x^2$ , and $n = 3$ . So, we use the substitution $v = y^{1-3} = y^{-2}$ . Then $\frac{dv}{dx} = -2y^{-3}\frac{dy}{dx}$ . Multiplying the original equation by $-2y^{-3}$ :

-2y^{-3}\frac{dy}{dx} - \frac{2}{x}y^{-2} = -2x^2

Substituting, we get:

\frac{dv}{dx} - \frac{2}{x}v = -2x^2

Now it's a linear equation in $v$ that you can solve. Remember to substitute back to find $y$ after solving for $v$ !

Tip: Always remember to substitute back to the original variable

y

after solving for

v

. Many students forget this crucial step!

Applications of Differential Equations

Differential equations are incredibly versatile. Here are a couple of important applications:

1. Growth and Decay

Many natural phenomena follow exponential growth or decay. For example, population growth (under ideal conditions) and radioactive decay can be modeled using these equations. The general formulas are:

Growth:

\frac{dP}{dt} = kP \implies P = P_0 e^{kt}

Decay:

\frac{dN}{dt} = -kN \implies N = N_0 e^{-kt}

Where $P$ is the population at time $t$ , $P_0$ is the initial population, $N$ is the amount of substance at time $t$ , $N_0$ is the initial amount, and $k$ is a constant. These come from directly solving the differential equations. For instance, for growth, separate the variables:

\frac{dP}{P} = k dt

Integrate both sides:

\int \frac{dP}{P} = \int k dt

ln(P) = kt + C

Exponentiate:

P = e^{kt + C} = e^C e^{kt}

Let $P_0 = e^C$ , which is the initial condition (at $t=0$ , $P=P_0$ ):

P = P_0 e^{kt}

The decay formula works similarly.

2. Newton's Law of Cooling

Newton's Law of Cooling describes how the temperature of an object changes when it's placed in a surrounding environment with a different temperature. The rate of change of the temperature is proportional to the temperature difference:

\frac{dT}{dt} = -k(T - T_s)

Where $T$ is the temperature of the object at time $t$ , $T_s$ is the surrounding temperature (assumed constant), and $k$ is a positive constant. Solving this equation gives:

T(t) = T_s + (T_0 - T_s)e^{-kt}

Where $T_0$ is the initial temperature of the object.

Let's see why this is true. Again, separate variables:

\frac{dT}{T - T_s} = -k dt

Integrate:

\int \frac{dT}{T - T_s} = \int -k dt

ln(T - T_s) = -kt + C

Exponentiate:

T - T_s = e^{-kt + C} = e^C e^{-kt}

Let $T_0$ be the initial temperature at $t=0$ . Then $e^C = T_0 - T_s$ . Thus:

T - T_s = (T_0 - T_s)e^{-kt}

T = T_s + (T_0 - T_s)e^{-kt}

Notice that as $t$ goes to infinity, $T$ approaches $T_s$ , which makes sense (the object eventually reaches the surrounding temperature).

Tip: When dealing with word problems, carefully identify the initial conditions and the constants involved. This will help you set up the differential equation correctly.

Common Mistake: Forgetting the negative sign in the decay equation or in Newton's Law of Cooling. Always remember that the rate of change is *negative* when the quantity is decreasing.

JEE Specific Tricks

For JEE Main, remember that many problems involving Bernoulli's equation can be solved quickly by recognizing the form and applying the substitution directly. Also, practice identifying the type of application (growth/decay, cooling) to set up the correct differential equation efficiently.

Here's a cool trick: if you see an equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$ , *immediately* consider the substitution $v = y^{1-n}$ . This can save you precious time during the exam.

Differential equations are a powerful tool. Mastering these concepts and practicing regularly will definitely give you an edge in the JEE Main exam! Keep practicing, and you'll conquer these problems in no time!