Euler's Theorem for Homogeneous Functions: A Secret Weapon for Differentiation
Master this powerful theorem that simplifies complex differentiation problems in JEE Advanced. Complete guide with proofs, applications, and practice problems.
Why Euler's Theorem is Your Secret Weapon
Euler's Theorem for Homogeneous Functions appears in 3-5 JEE Advanced questions every year, yet many students overlook its power. Mastering this theorem will give you:
- One-step solutions to complex differentiation problems
- Time savings of 2-3 minutes per question
- Elimination of messy calculations in multivariable calculus
- Confidence boost in solving "impossible-looking" problems
Euler's Theorem Statement
Definition:
A function $f(x, y)$ is homogeneous of degree $n$ if:
$f(tx, ty) = t^n f(x, y)$ for all $t > 0$
Euler's Theorem:
If $f(x, y)$ is homogeneous of degree $n$, then:
$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(x, y)$
Quick Proof:
Step 1: Start with $f(tx, ty) = t^n f(x, y)$
Step 2: Differentiate both sides with respect to $t$:
$x \frac{\partial f}{\partial (tx)} + y \frac{\partial f}{\partial (ty)} = n t^{n-1} f(x, y)$
Step 3: Set $t = 1$:
$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(x, y)$
Q.E.D.
Application 1: Quick Verification of Homogeneity
Verify that $f(x, y) = x^3 + 3x^2y + y^3$ is homogeneous and find its degree.
Solution Approach:
Step 1: Check $f(tx, ty)$:
$f(tx, ty) = (tx)^3 + 3(tx)^2(ty) + (ty)^3$
$= t^3x^3 + 3t^3x^2y + t^3y^3 = t^3(x^3 + 3x^2y + y^3)$
Step 2: Since $f(tx, ty) = t^3 f(x, y)$, degree $n = 3$
Step 3: Verify Euler's theorem:
$\frac{\partial f}{\partial x} = 3x^2 + 6xy$, $\frac{\partial f}{\partial y} = 3x^2 + 3y^2$
$x(3x^2 + 6xy) + y(3x^2 + 3y^2) = 3x^3 + 6x^2y + 3x^2y + 3y^3$
$= 3(x^3 + 3x^2y + y^3) = 3f(x, y)$ ✓
Application 2: Finding Partial Derivatives Quickly
If $u = \tan^{-1}\left(\frac{x^3 + y^3}{x - y}\right)$, prove that $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \sin 2u$
Solution Approach:
Step 1: Let $f(x, y) = \frac{x^3 + y^3}{x - y}$
Step 2: Check homogeneity:
$f(tx, ty) = \frac{t^3x^3 + t^3y^3}{tx - ty} = t^2 \frac{x^3 + y^3}{x - y} = t^2 f(x, y)$
Step 3: So $f$ is homogeneous of degree 2
Step 4: By Euler's theorem:
$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 2f$
Step 5: Since $u = \tan^{-1} f$, we have:
$\frac{\partial u}{\partial x} = \frac{1}{1+f^2} \frac{\partial f}{\partial x}$, $\frac{\partial u}{\partial y} = \frac{1}{1+f^2} \frac{\partial f}{\partial y}$
Step 6: Multiply and add:
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{1}{1+f^2}(x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y}) = \frac{2f}{1+f^2}$
Step 7: But $\frac{2f}{1+f^2} = \sin 2u$ since $f = \tan u$
Application 3: Solving Partial Differential Equations
If $u = f\left(\frac{y}{x}\right)$, prove that $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0$
Solution Approach:
Step 1: Recognize that $u = f\left(\frac{y}{x}\right)$ is homogeneous of degree 0
Step 2: Check: $u(tx, ty) = f\left(\frac{ty}{tx}\right) = f\left(\frac{y}{x}\right) = t^0 u(x, y)$
Step 3: By Euler's theorem with $n = 0$:
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0 \cdot u = 0$
Step 4: Direct verification:
Let $v = \frac{y}{x}$, then $\frac{\partial u}{\partial x} = f'(v) \cdot (-\frac{y}{x^2})$
$\frac{\partial u}{\partial y} = f'(v) \cdot (\frac{1}{x})$
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x(-\frac{y}{x^2}f'(v)) + y(\frac{1}{x}f'(v)) = 0$ ✓
🚀 Euler's Theorem Problem-Solving Framework
Identification Tips:
- Look for functions where each term has same total power
- Check if $f(tx, ty) = t^n f(x, y)$ pattern exists
- Functions of $\frac{y}{x}$ or $\frac{x}{y}$ are degree 0
- Sum of homogeneous functions has degree of highest term
Application Strategies:
- Use theorem to verify your differentiation
- Apply for quick elimination in multiple choice
- Combine with chain rule for complex functions
- Remember the extension to three variables
Applications 4-5 Available in Full Version
Includes extension to three variables and advanced problem-solving techniques
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. If $u = \frac{x^2 + y^2}{x + y}$, find $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$
2. Verify Euler's theorem for $f(x, y) = \frac{x^4 - y^4}{x^2 + y^2}$
3. If $z = x^n f\left(\frac{y}{x}\right)$, prove that $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = nz$
Extension to Three Variables
Theorem:
If $f(x, y, z)$ is homogeneous of degree $n$, then:
$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} = n f(x, y, z)$
Example:
If $u = \frac{x^2 + y^2 + z^2}{xy + yz + zx}$, find $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z}$
Solution: Check that $u$ is homogeneous of degree 0, so the expression equals 0.
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