Problem 70 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Solution of the differential equation \(2 y \sin x \frac{d y}{d x}=\) \(2 \sin x \cos x-y^{2} \cos x\) satisfying \(y\left(\frac{\pi}{2}\right)=1\) is given by (A) \(y^{2}=\sin x\) (B) \(y=\sin ^{2} x\) (C) \(y^{2}=\cos x+1\) (D) \(y^{2} \sin x=4 \cos ^{2} x\)

Step-by-Step Solution

Answer

(A) \(y^{2}=\sin x\)

Step 1: Analysis of the Differential Equation

We start by rewriting the differential equation as given: \(2 y \sin x \frac{dy}{dx} = 2 \sin x \cos x - y^2 \cos x\). We can separate the terms involving \(y\) and \(x\) if possible.

Step 2: Rearrange the Equation

Rearranging the equation, we get: \(2 y \sin x \frac{dy}{dx} + y^2 \cos x = 2 \sin x \cos x\). We can solve this equation using methods for separable or linear differential equations.

Step 3: Substitute to Simplify

Observe that since the equation involves \(y\) and \(y^2\), we can let \(v = y^2\) such that \(y \frac{dy}{dx} = \frac{1}{2} \frac{dv}{dx}\). Substituting, get \(\sin x \frac{dv}{dx} + v \cos x = 2 \sin x \cos x\).

Step 4: Solve the Linear Equation

This new equation can be tackled as a linear first-order differential equation of the form \(\frac{dv}{dx} + \frac{v \cos x}{\sin x} = 2 \cos x\). Use integrating factors or substitution to solve this.

Step 5: Use Integrating Factor

The integrating factor is \(e^{\int \cot x \, dx} = \sin x\), thus multiplying through by this factor yields \(\sin x \frac{dv}{dx} + v \cos x = 2 \sin x \cos x\) simplifies to \(\frac{d}{dx}(v \sin x) = 2 \sin x \cos x\).

Step 6: Integrate Both Sides

Integrate the equation: \(\int \frac{d}{dx}(v \sin x) \, dx = \int 2 \sin x \cos x \, dx\), giving \(v \sin x = \sin^2 x + C\). Solve \(2 \sin x \cos x\) as \(\sin(2x)\), and integrate accordingly.

Step 7: Apply Initial Conditions

Using the initial condition \(y\left(\frac{\pi}{2}\right)=1\), then \(v = y^2\), we have \(v \sin\left(\frac{\pi}{2}\right) = 1^2 = 1\). Substitute \(x = \frac{\pi}{2}\) and solve for \(C\).

Step 8: Conclusion

Setting \(x = \frac{\pi}{2}\) yields \(v = 1\) and hence \(\sin^2 x + C = 1\). The integration leads to \(v = \sin x\) with \(v = y^2\). Therefore, the solution is \(y^2 = \sin x\), which is option (A).

Key Concepts

Understanding Initial Conditions in Differential Equations

Initial conditions play a crucial role in solving differential equations. They're used to determine the specific solution to a differential equation from a family of potential solutions. Essentially, they provide the necessary information to pinpoint the exact path your function takes among many possibilities. In our given exercise, the initial condition is expressed as \( y\left(\frac{\pi}{2}\right) = 1 \). This means that at \( x = \frac{\pi}{2} \), the function \( y \) equals 1.

• Initial conditions provide specific point values that help solve the equation.
• They help convert a general solution into a unique solution.

For instance, in the final steps of solving our differential equation, we use these initial conditions to solve for the constant \( C \) that appears after integration. Applying these conditions allows us to substitute and determine that the value of \( v = y^2 \) when \( x = \frac{\pi}{2} \) must equate to a certain value gathered from our exercise, ensuring that our result satisfies the entire equation fully.

Exploring the Integrating Factor Technique

The integrating factor is a powerful method used to solve linear first-order differential equations. This technique simplifies the differential equation, making it easier to find a solution. In our exercise, the differential equation, after a clever substitution, takes the form \( \frac{dv}{dx} + \frac{v \cos x}{\sin x} = 2 \cos x \). The integrating factor \( IF( x ) = e^{\int \cot x \, dx} = \sin x \) is used to transform this equation.

• Multiply each term by the integrating factor to make the equation easier to solve.
• Transforms the left hand side into an exact derivative.

By applying this factor, the differential equation evolves into \( \frac{d}{dx}(v \sin x) = 2 \sin x \cos x \), allowing us to integrate both sides easily. This transformation clears the path to finding an exact solution by simply integrating, because the left side becomes a total derivative, making it straightforward to solve.

Decoding Linear First-Order Differential Equations

Linear first-order differential equations are an important class of equations in calculus and have the general form \( \frac{dy}{dx} + P(x)y = Q(x) \). They are called 'linear' because both the dependent variable \( y \) and its derivative \( \frac{dy}{dx} \) appear to the first power (i.e., linearly). The equation in our step-by-step solution is modified into a linear form using a substitution, illustrating this problem's structure.

• Create a standard form through rearrangement.
• Once in linear form, use the integrating factor for solving.

In our case, after substituting \( v = y^2 \), the differential equation becomes \( \frac{dv}{dx} + \frac{v \cos x}{\sin x} = 2 \cos x \), showcasing its linear nature. The systematic approach of recognizing and transforming into the linear form allows us to exploit effective solution techniques like integrating factors, as demonstrated. Recognizing these patterns is key to tackling wider classes of problems within differential equations.