Problem 30 - Calculus for JEE ADVANCED

Question

\(\left(x^{2}+x y\right) d y=\left(x^{2}+y^{2}\right) d x\) or \(\frac{d y}{d x}=\frac{x^{2}+y^{2}}{x^{2}+x y}\)

Step-by-Step Solution

Answer

Use substitution \(v = \frac{y}{x}\) and separate variables to solve.

Step 1: Rearrange the Equation

Start by rewriting the given differential equation in a separable form. Divide both sides by \(x^2 + xy\) to obtain \(\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}\).

Step 2: Check for Exactness

Determine if the equation is exact by examining the functions \(M(x,y) = x^2 + y^2\) and \(N(x,y) = x^2 + xy\). Check if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). If: \( \frac{\partial M}{\partial y} = 2y\) and \( \frac{\partial N}{\partial x} = 2x+y\), the equation is not exact.

Step 3: Find an Integrating Factor or Substitute

Since the equation is not exact, look for an integrating factor or appropriate substitution. Notice that the substitution \(v = \frac{y}{x}\) can help simplify since the differential involves both \(x\) and \(y\).

Step 4: Apply Substitution

Use the substitution \(y = vx\), which implies \(dy = vdx + xdv\). Substitute \(dy\) and \(y\) in the differential equation: \(x^2(1+v^2)dx + (x^2 + x^2v)dx + x^2dv = 0\).

Step 5: Separate Variables with Substitution

After substitution, rearrange the terms to isolate \(dv\) and \(dx\): \((1+v^2+v)dx + xdv = 0\). This further simplifies to \(dx/x = -dv/(1+v+v^2)\).

Step 6: Integrate Both Sides

Integrate both sides separately: \(\int \frac{1}{x} dx = -\int \frac{1}{1+v+v^2} dv\). Solve these integrals to get \(\ln|x| = -\int \frac{1}{1+v+v^2} dv + C\), where \(C\) is the integration constant.

Step 7: Solve the Integral and Back-Substitute

Evaluate the integral on the right. Since \(1 + v + v^2\) is a quadratic, use partial fractions or a trigonometric substitution if necessary. After solving, back-substitute to replace \(v\) with \(y/x\).

Step 8: Combine and Simplify

Combine the results from the integration and simplify. Substitute any constants back to form the implicit solution. The final expression will involve \(x\) and \(y\).

Key Concepts

Exactness

In solving differential equations, checking for exactness is an important step. An exact differential equation is one where a potential function exists. This function is indirectly differentiated with respect to both variables involved.
To determine if an equation is exact, we look for two functions, usually denoted as \( M(x, y) \) and \( N(x, y) \). In our given exercise, these are \( M(x, y) = x^2 + y^2 \) and \( N(x, y) = x^2 + xy \).

• We calculate \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).
• If \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), then the equation is exact.
• If not, like in our exercise where \( \frac{\partial M}{\partial y} = 2y \) and \( \frac{\partial N}{\partial x} = 2x + y \), the equation is not exact.

Recognizing non-exact equations helps in knowing that an integrating factor or substitution is required for solving.

Integrating Factor

When a differential equation is not exact, finding an integrating factor is a method to convert it into an exact equation. An integrating factor is a function that, when multiplied by the original equation, makes it exact.
The method involves identifying a factor \( \mu(x, y) \), such that when multiplied by both \( M(x, y) \) and \( N(x, y) \):

• The equation becomes exact.
• This often requires checking specific conditions or symmetries in the problem.

For our specific problem, we were advised to try a substitution instead of directly finding an integrating factor. But if the substitution method were not applicable, an integrating factor could be a suitable alternative.

Variable Substitution

In some cases, like our specific differential equation, a change of variables can simplify the equation. Substituting variables can transform a non-linear differential equation into a simpler form.
In the exercise, variable substitution was key:

• We used \( v = \frac{y}{x} \), simplifying the equation by expressing \( y = vx \) and consequently \( dy = vdx + x dv \).
• This substitution simplified the equation into a form where separating variables is feasible.

This technique can often align multiple variables into a single variable problem, making integration straightforward.

Separation of Variables

Separation of variables is a technique for solving differential equations by isolating the variables on opposite sides of the equation. This technique is often used after applying a suitable substitution.
In our exercise, after substituting \( v = \frac{y}{x} \), we derived the equation \( dx/x = -dv/(1+v+v^2) \). Here’s how it's tackled:

• By integrating both sides separately, we solve for \( x \) in terms of \( v \).
• Integration involves finding the antiderivative for both \( \int \frac{1}{x} dx \) and the more complex \( \int \frac{1}{1+v+v^2} dv \).

This concept has put us in a position to solve our differential equation by handling simpler integrals, eventually enabling back-substitution to get an expression involving the original variables \( x \) and \( y \).