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Euler's Theorem of Homogeneous Functions Euler's Theorem of Homogeneous Functions Euler's theory of equivalent functions clarifies a key concept in statistical analysis, showing the relationship between equivalent functions. Discover deeper insights into the scaling behavior of these functions and their impact on mathematical models, optimization, and applications. Unpack the basic principles of Euler's theorem and dive clearly and deeply into the fascinating world of identical functions. If \(z\) be a homogeneous function of \(x, y\) of order \(n\), then \(x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz\) all Proof: Since \(z\) is a homogeneous function of \(x, y\) of order \(n\). \(z=x^{n}f(\frac{y}{x}) \dots (1)\) \(\frac{\partial z}{\partial x}=n~x^{n-1}f(\frac{y}{x})+x^{n}.f^{\prime}(\frac{y}{x}).(-\frac{y}{x^{2}}).\) or ...

Charpit’s Method Charpit’s Method or General Method to Solve Non-Linear Partial Differential Equation Charpit’s Method: It is a well-known mathematical technique known as “Method of Characteristics”. The method is widely used for solving first-order or quasi-linear PDE’s. Explanation of Method: Let the given differential equation be: Different Types of Solutions in PDE \[ f(x, y, z, p, q) = 0 \quad \text{(i)} \quad \text{where} \quad z = z(x, y) \] We know \( dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy \). [Therefore, z=z(x,y)] \( dz = p \, dx + q \, dy \quad \text{(ii)} \). Now we shall find another relation \( F(x, y, z, p, q) = 0 \quad \text{(iii)} \) so that on solving equations (i) and (iii) for \( p \) and \( q \) and putting these values in (ii), equation (ii) becomes integrable, and this integral gives the complete solution of (i) (integral). For finding \( F \), differentiate (i) with resp...

In Partial Differential Equation questions are solving to some important forms Lagrange's Linear Equation, Charpit's Method and etc. Here the concept of solutions are very vital to understand for solving the P.D Equation. Types of Solutions There are four types of solutions are listed below - Complete Solution Particular Solution Singular Solution General Solution Complete Solution When a partial differential equation f(x,y,z,p,q) = 0      ...(i) where, z is the function of independent variables x and  y and f   be the function of dependent variables. Here we obtain a relation g(x,t,z,a,b) = 0 there are many arbitary constants as a number of independent variables, then g(x,y,z,p,q) = 0 is called the complete solution or integral of eq( i ) for example, z = ax + by +f(a,b) is the complete solution of z = px +qy +f(p,q) Particular Solution First of all we have a differential equatio...

Linear Independence and Wronskian Show that the following pair of functions are linearly independent yet their Wronskian vanishes on the given interval $$f_{1}=\begin{cases}x^{2},x\ge0\\ 0,x Sol. The given functions are $$f_{1}=\begin{cases}x^{2},x\ge0\\ 0,x We want to prove that $$f_{1}$$, $$f_{2}$$ are L.I. For this purpose, we shall show that ...(1) if $$c_{1}f_{1}+c_{2}f_{2}=0$$ for all $$x\in\mathbb{R}$$, then each of $$c_{1}$$, $$c_{2}$$ is zero. For this purpose, we use $$x=-1.1$$. Now $$f_{1}(1)=1$$, $$f_{2}(1)=0$$, and $$f_{1}(-1)=0$$, $$f_{2}(-1)=1$$. Substituting these values in (1), we get, $$c_{1}\cdot1+c_{2}\cdot0=0\Rightarrow c_{1}=0$$ and $$c_{1}\cdot0+c_{2}\cdot1=0\Rightarrow c_{2}\rightarrow0$$ We have $$c_{1}f_{1}+c_{2}f_{2}=0\Rightarrow c_{1\frac{3}{3}}=c_{2}=0$$ Hence $$f_{1}$$, $$f_{2}$$ are linearly independent on the given interval. Now, we find the Wronskian of the given functions. Two cases arise. Case I. When $$x\...

  Let          f1(x) = x,     f2(x) = x^3,     f3(x) = x^4                  f'1(x) = 1,    f'2(x) = 3x^2  f'(x) = 4x^3                  f"1(x) = 0,   f"2(x) = 6x,    f"3(x) = 12x^2 W( f 1, f 2, f 3) =                     =  Taking x common form 1st Row                =                =  Now,                                         = x(24x^4 - 18x^4)                                        = 6x^5 which is not equal to o Therefore...

Th. Prove that if the Wronskian of the functions \(f_{1}\, f_{2}\, \ldots f_{n}\) over an Interval I is non-zero, then the functions are linearly independent over I. Proof Proof: Consider the relation \[ c_{1}f_{1} + c_{2}f_{2} + \ldots + c_{n}f_{n} = 0 ...(1)\] where \( z_{1}, c_{2}, \ldots, c_{n} \) are constants. Differentiating (1) successively \( n-1 \) times with respect to \( x \), we get, \[ c_{1}f'_{1} +c_{2}f'_{2} +c_{3}f'_{3} + \ldots + c_{n} {f_{n}}^{\prime} = 0 ...(2)\] \[ c_{1}f''_{1} + c_{2}f''_{2} + \ldots + c_{n}f''_n = 0 ...(3)\] \[ c_{1}{f_{1}}^{(n-1)} + c_{2}{f_{2}}^{(n-1)} + \ldots + c_{n}{f_{n}}^{(n-1)} = 0 ...(n)\] Here The Vidoe in Youtube These \( n \) equations can be written as \[ \begin{bmatrix} f_{1} & f_{2} & \ldots & f_{n} \\ f'_{1} & f'_{2} & \ldots & f'_{n} \\ f"_{1} & f"_...