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

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

The solution is (A) \(y^2 = \sin x\).

Step 1: Rearrange the given differential equation

We start by rearranging the given differential equation \(2 y \sin x \frac{dy}{dx} = 2 \sin x \cos x - y^2 \cos x\). Notice that we can factor out \(\sin x\) from both terms on the right-hand side: \( \sin x (2 \cos x - y^2 / y) \). This can be rewritten as \( \sin x (2 \cos x - y)\). This means the equation can be rearranged to \(2 y \frac{dy}{dx} = 2 \cos x - y^2 / y\).

Step 2: Simplify and integrate both sides

Divide through the equation carefully and separate variables if possible. We attempt to move terms implying the derivative \(dy/dx\) to one side.\(2y \frac{dy}{dx} = \cos x (2 - y^2)\). Upon separation and simplification, this becomes: \( \frac{2y}{2y - y^2} \frac{dy}{dx} = \cos x\). Integrate both sides with respect to \(x\).

Step 3: Complete integration

The integration on the left requires careful handling: \(\int \frac{2y}{2y - y^2} dy = \int \cos x dx\). Solve the integral to compute both sides. This results in finding \(y(x)\) with integration constant determined by initial conditions.

Step 4: Apply the initial condition

Utilizing the given initial condition \(y(\frac{\pi}{2}) = 1\), insert \(x = \frac{\pi}{2}\) and \(y = 1\) in the integral solutions to resolve for the constant determined in the previous integration.

Step 5: Verify the general solution matches options provided

After applying the initial condition, determine which of the given options matches the form of the solution found from integration. Using replacement and simplification methods verify solution compatibility focused on the expression \(y^2 = \sin x\).

Key Concepts

Initial Conditions

Differential equations often include specific values called initial conditions that help us find unique solutions. An initial condition provides a point which the solution curve must go through. This greatly narrows down the possibilities for a solution to the differential equation.
For example, if our solution is a family of curves, the initial condition will allow us to pinpoint the exact curve that satisfies both the differential equation and the initial condition. In our current exercise, the initial condition given is \(y\left(\frac{\pi}{2}\right) = 1\).
This means our solution curve must pass through the point \((x, y) = (\frac{\pi}{2}, 1)\). This helps to determine an arbitrary constant resulting from the integration process.

• It ensures that out of many possible solution curves, we find the one that passes through our specified point.
• It leads to a particular solution, distinguishing it from the general solution.

As a simple analogy, consider a road map where the road is looping across hills (differential equation). The initial condition is akin to a specific checkpoint on that road that your vehicle must pass; this checkpoint ensures you're on the correct section of the road for your journey.

Separation of Variables

The technique of separation of variables assists in solving a differential equation by rewriting it so that each variable and its derivative are placed on opposite sides of the equation. This simplifies the problem substantially by allowing the integration of both sides independently.
In our problem, the equation was rearranged so that all terms involving \(y\) and \(dy\) are on one side, while those involving \(x\) and \(dx\) are on the other. We achieved this through rearrangement and simplification:
This yielded: \(\frac{2y}{2y - y^2} \frac{dy}{dx} = \cos x\).
The ease of having one function per variable on each side cannot be overstated, as it transforms a daunting process into a manageable task, easily expressed in terms of integrals.

• It's often more straightforward than other methods in solving differential equations.
• By directly integrating both sides, you effectively boil down the problem to solving two simpler problems.

Harnessing this simplification effectively decouples the relationship between \(x\) and \(y\) in your equation, making it easier to solve than keeping them entangled.

Integration of Differential Equations

Once the variables are separated, each side of the differential equation can be integrated with respect to its own variable. Integrating differential equations converts the problem from dealing with rates of change (derivatives) to dealing with quantities (integrals).
In our scenario, after separating variables, you integrate:
\[ \int \frac{2y}{2y - y^2} \,dy = \int \cos x \,dx \] By integrating the left side with respect to \(y\), and the right side with respect to \(x\), we aim to find a function \(y(x)\).
This functions as a bridge over which the rate-based description (differential equation) transforms into a quantity-based description (integral solution).

• The integration gives the general solution to the differential equation in terms of an unknown constant.
• This constant can be determined when an initial condition is applied.

Thus, integrating allows us to untangle the original rate relation, offering a clearer understanding or expression of the entire process being modeled, as well as predicting future values given initial conditions.