Problem 12 - Calculus for JEE ADVANCED

Question

\(\frac{1}{y+1} d y=-\frac{\cos x}{2+\sin x} d x\)

Step-by-Step Solution

Answer

\(y = \frac{C}{|2 + \sin x|} - 1\)."

Step 1: Separate Variables

Start by separating the variables so that all terms involving \(y\) are on one side of the equation and all terms involving \(x\) are on the other side. The given equation is \(\frac{1}{y+1} dy = -\frac{\cos x}{2 + \sin x} dx\). This equation is already separated with \(y\) terms on the left and \(x\) terms on the right.

Step 2: Integrate Both Sides

Integrate both sides of the equation separately. The left side becomes \(\int \frac{1}{y+1} \, dy\), which integrates to \(\ln|y+1| + C_1\). The right side becomes \(-\int \frac{\cos x}{2 + \sin x} \, dx\). To integrate the right side, use the substitution \(u = 2 + \sin x\), yielding \(du = \cos x \, dx\). This simplifies the integral to \(-\int \frac{1}{u} \, du\), which results in \(-\ln|u| + C_2\). Substituting back for \(u\), you get \(-\ln|2 + \sin x| + C_2\).

Step 3: Combine the Integrated Expressions

After integrating, combine the expressions for both sides:\[ \ln|y+1| = -\ln|2 + \sin x| + C \]where \(C = C_2 - C_1\).

Step 4: Solve For y

Exponentiate both sides to solve for \(y\):\[ y+1 = Ce^{-\ln|2+\sin x|} = \frac{C}{|2 + \sin x|} \]So, \(y = \frac{C}{|2 + \sin x|} - 1\).

Key Concepts

Variable Separation

To solve a differential equation like the one given in the original exercise, a common technique is variable separation. This approach involves manipulating the equation so that all terms containing the dependent variable, in this case, \( y \), are on one side of the equation while all terms involving the independent variable, \( x \), are on the other. Variable separation essentially prepares the equation for integration. In our given exercise:

• We have \( \frac{1}{y+1} dy \) and \( -\frac{\cos x}{2 + \sin x} dx \), meaning the equation is already separated.
• The left side entirely involves \( y \) and the right side involves \( x \).

By separating variables, we simplify the integration process, making it straightforward to address each variable independently. This is why variable separation is often one of the first steps in solving differential equations with separable variables.

Integration Techniques

Once the equation is separated, the next task is to integrate both sides. Integration techniques provide the tools to find the antiderivatives, which are crucial in solving differential equations. For the left side of our equation:

• The integral \( \int \frac{1}{y+1} \, dy \) is straightforward, yielding \( \ln|y+1| + C_1 \). This is a standard integral resulting from the natural logarithmic function.

The right side involves a trigonometric substitution:

• Integration might require substitution or techniques such as partial fraction decomposition, integration by parts, etc., but here we use substitution.

Understanding various integration techniques allows you to tackle a diverse set of integrals, making it invaluable in solving differential equations.

Integration by Substitution

A powerful tool in integration is substitution, which simplifies integrals by changing variables. This technique is particularly useful when the integral contains complex functions that defy simpler methods. On the right side of our original exercise equation, the integral \( -\int \frac{\cos x}{2 + \sin x} \, dx \) requires substitution:

• We let \( u = 2 + \sin x \). This changes the differential \( du = \, \cos x \, dx \), matching perfectly with the integrand's \( \cos x \, dx \).
• The integral transforms into \( -\int \frac{1}{u} \, du \), which simplifies to \(-\ln|u| + C_2 \).

After solving, we substitute back \( u = 2 + \sin x \) to complete the process. Integration by substitution is a versatile technique, easing the path to solution for integrals involving functions nested within other functions.