Problem 52 - Calculus for JEE ADVANCED

Question

\(f^{\prime}(x)-\frac{2 x(x+1)}{x+1} f(x)=\frac{e^{x^{2}}}{(x+1)^{2}}\)

Step-by-Step Solution

Answer

The solution is \(f(x) = e^{x^2} \left(-\frac{1}{x+1} + C\right)\).

Step 1: Identify the Type of Differential Equation

The given differential equation is \(f^{\prime}(x) - \frac{2x(x+1)}{x+1}f(x) = \frac{e^{x^2}}{(x+1)^2}\). It's a first-order linear differential equation of the form \(y' + P(x)y = Q(x)\). Here, \(P(x) = -2x\) and \(Q(x) = \frac{e^{x^2}}{(x+1)^2}\).

Step 2: Find the Integrating Factor

The integrating factor \(\mu(x)\) is obtained by \(\mu(x) = e^{\int P(x) \, dx} = e^{\int -2x \, dx}\). The integral \(\int -2x \, dx = -x^2\), so \(\mu(x) = e^{-x^2}\).

Step 3: Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \(e^{-x^2}\):\[ e^{-x^2}f'(x) - 2x e^{-x^2} f(x) = \frac{e^{x^2}}{(x+1)^2} e^{-x^2} \]Simplifying, we have:\[ e^{-x^2}f'(x) - 2x e^{-x^2} f(x) = \frac{1}{(x+1)^2} \]

Step 4: Recognize the Left Hand Side as a Derivative

The left-hand side \(e^{-x^2}f'(x) - 2xe^{-x^2}f(x)\) can be recognized as the derivative of \((e^{-x^2}f(x))\) with respect to \(x\). This is due to the product rule of derivatives. Thus, the equation is:\[ \frac{d}{dx}(e^{-x^2} f(x)) = \frac{1}{(x+1)^2} \]

Step 5: Integrate Both Sides

Integrate the equation \(\frac{d}{dx}(e^{-x^2} f(x)) = \frac{1}{(x+1)^2}\) with respect to \(x\):\[ e^{-x^2} f(x) = \int \frac{1}{(x+1)^2} \, dx \]The integral \(\int \frac{1}{(x+1)^2} \, dx = -\frac{1}{x+1} + C\), where \(C\) is the constant of integration.

Step 6: Solve for \(f(x)\)

From the integrated equation, solve for \(f(x)\):\[ e^{-x^2} f(x) = -\frac{1}{x+1} + C \]Multiply both sides by \(e^{x^2}\) to solve for \(f(x)\):\[ f(x) = e^{x^2} \left(-\frac{1}{x+1} + C\right) \]

Key Concepts

Integrating Factor

To solve first-order linear differential equations, the integrating factor is an essential tool. This method involves a function, denoted as \( \mu(x) \), which helps simplify and solve the differential equation, making the process much more straightforward.

In our specific equation, the integrating factor is calculated from the part of the equation that is of the form \( y' + P(x)y \). Here, \( P(x) = -2x \). The integrating factor is found using the formula \( \mu(x) = e^{\int P(x) \, dx} \). Thus, the integral \( \int -2x \, dx \) gives us \( -x^2 \), leading to the integrating factor being \( \mu(x) = e^{-x^2} \).

Once determined, we multiply the entire equation by this integrating factor. This transforms the left side of the equation into the derivative of the product \( \mu(x) \times f(x) \), which helps in further simplifying the entire problem.

Product Rule

The product rule is a fundamental principle in calculus that helps us differentiate products of two functions. Sometimes, in solving differential equations, particularly when using an integrating factor, identifying a derivative on the left side of an equation can be key to simplifying it.

For our given problem, after multiplying by the integrating factor, the left-hand side simplifies to a format represented by the derivative of a product. Specifically, the expression \( e^{-x^2}f'(x) - 2x e^{-x^2}f(x) \) is identified as the derivative of \( e^{-x^2}f(x) \). This is a direct application of the product rule, expressing that the derivative of \( uv \) is \( u'v + uv' \).

• \( u = e^{-x^2} \)
• \( v = f(x) \)

Recognizing this identity allows us to rewrite the differential equation in a simpler form that is ready to be integrated easily.

Constant of Integration

The constant of integration is a crucial component when dealing with indefinite integrals. Whenever you integrate a function, a constant is added to represent all possible vertical shifts of the antiderivative.

When integrating both sides of the simplified differential equation \( \frac{d}{dx}(e^{-x^2} f(x)) = \frac{1}{(x+1)^2} \), we find the antiderivative on the right as \( -\frac{1}{x+1} + C \), where \( C \) is the constant of integration.

• This constant represents all potential solutions to the differential equation.
• It's essential in initial and boundary value problems where specific conditions are given.

In our solution, \( C \) allows us to express the general solution of the differential equation, thereby representing an entire family of potential solutions depending on the initial conditions.

Solving Differential Equations

Solving differential equations can initially seem complex, but breaking it down into manageable steps simplifies the process significantly.

To solve the first-order linear differential equation provided:

• First, identify the structure in the standard form \( y' + P(x)y = Q(x) \).
• Calculate the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \).
• Multiply the differential equation through by the integrating factor to utilize the product rule and express it as a total derivative.
• Integrate both sides of the resulting equation to find an expression for the unknown function, \( f(x) \).
• Don’t forget to include the constant of integration when you integrate.

After following these steps, obtain the solution \( f(x) = e^{x^2} \left(-\frac{1}{x+1} + C\right) \). This provides the general solution for the desired function. Remember that with practice, recognizing these patterns and steps becomes quicker and more intuitive.