Problem 81 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Solution of the differential equation \(y d x+\left(x+x^{2} y\right) d y=0\) is (a) \(\log y=C x\) (b) \(-\frac{1}{x y}+\log y=C\) (c) \(\frac{1}{x y}+\log y=C\) (d) \(-\frac{1}{x y}=C\)

Step-by-Step Solution

Answer

(b) \(-\frac{1}{xy} + \log y = C\) matches the solution obtained.

Step 1: Identify the Differential Equation Type

The given differential equation is \( y \, dx + (x + x^2 y) \, dy = 0 \). This can be rearranged to \( y \, dx + (x + x^2 y) \, dy = 0 \) or rewritten as \( M(x, y) \, dx + N(x, y) \, dy = 0 \), where \( M(x, y) = y \) and \( N(x, y) = x + x^2 y \). This format suggests that it could be an exact differential equation.

Step 2: Check for Exactness

To determine if the differential equation is exact, we must check if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Compute each: \( \frac{\partial M}{\partial y} = 1 \) and \( \frac{\partial N}{\partial x} = 1 + 2xy \). Since \( \frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x} \), the differential equation is not exact.

Step 3: Consider a Substitution

Since the equation is not exact, we try to simplify it by substitution. Recognize that, if rewritten, could form the role compatible with a Bernoulli's equation. Let's attempt a substitution like \( v = xy \), where \( dy = \frac{dv}{x} - \frac{v}{x^2}dx \). Substitute into the equation and simplify.

Step 4: Substitute and Rearrange

Substitute \( v = xy \), then \( y = \frac{v}{x} \) and \( dy = \frac{vdx - xdv}{x^2} \). Substitute these into the original equation, considering the modification: \( \Rightarrow y \, dx = \frac{v}{x} \, dx \). Substitute again and simplify to obtain a potentially solveable form.

Step 5: Solve the Simplified Equation

After substitution and rearranging, solve the resulting form of the differential equation. Eventually, integrate if necessary to solve for \( y \) in terms of \( x \).

Step 6: Verify Solution Against Options

Compare your solution from Step 5 with the given solutions in the options (a) \( \log y = Cx \), (b) \( -\frac{1}{xy} + \log y = C \), (c) \( \frac{1}{xy} + \log y = C \), and (d) \( -\frac{1}{xy} = C \). Check which solution fits or satisfies the obtained equation.

Key Concepts

Exact Differential Equation

An exact differential equation is one where you can integrate both parts seamlessly to find the solution. It follows the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \). For a differential equation to be exact, we need to check a specific condition involving derivatives:

• The partial derivative of \( M(x, y) \) with respect to \( y \) should equal the partial derivative of \( N(x, y) \) with respect to \( x \).

In the given problem, \( M(x, y) = y \) and \( N(x, y) = x + x^2 y \). After calculating partial derivatives, we found that \( \frac{\partial M}{\partial y} = 1 \) is not equal to \( \frac{\partial N}{\partial x} = 1 + 2xy \).
Thus, the equation \( y \, dx + (x + x^2 y) \, dy = 0 \) is not exact. To handle non-exact equations, we often look for substitutions or integrating factors to transform them into exact ones.

Bernoulli's Equation

Named after the Swiss mathematician Jacob Bernoulli, Bernoulli's equation is a special type of first-order differential equation. It typically takes the form: \[\frac{dy}{dx} + P(x)y = Q(x)y^n\]Here, \( n \) must be any real number except 0 or 1. Even though it varies slightly from the linear form, it can be transformed into a linear equation using a substitution method.
When we examined the given equation, recognizing potential similarities with Bernoulli's form encouraged us to attempt a substitution. We used \( v = xy \) to set up a relationship that facilitated the transformation.
This transformation helps align Bernoulli's equation to a solvable form, allowing for further steps to solve it.

Substitution Method

The substitution method is a valuable technique in solving differential equations that aren't easily solvable in their original form. It involves changing variables to simplify the equation. The original equation \( y \, dx + (x + x^2 y) \, dy = 0 \) seemed complex and non-exact. To simplify, we defined a substitution like \( v = xy \).
Through substitution, transformations convert terms involving \( y \) into expressions of \( v \). As a result, the equation takes on a new form that may be more aligned with known solvable types. After substituting, we ended up with new terms such as \( y = \frac{v}{x} \) and derived differential relationships to replace \( dy \). In creating a cohesion of like terms, substitution plays a crucial role in solving complex differential equations by streamlining them into simpler linear or more manageable forms.

Partial Derivatives

Partial derivatives are instrumental in understanding how functions change as their input variables alter. They measure how a function's output varies as each input changes, considering all others as constant.
In the context of our differential equation, partial derivatives helped us assess exactness by providing a way to relate \( M(x, y) \) and \( N(x, y) \) functions:

• We calculated \( \frac{\partial M}{\partial y} = 1 \).
• We calculated \( \frac{\partial N}{\partial x} = 1 + 2xy \).

By comparing the outcomes of these partial derivatives, we determined the non-exactness. Understanding how to use and compute partial derivatives is crucial when dealing with multi-variable functions and assessing exactness in differential equations.