Mathquery

Mathematics is thuniverse of knowledge. From starting of universe it is a secret of all objects of this universe. Every ending has a starting. Differential Equations is the part of  universe of Mathematics.              Differential equations are divided into two parts those are Ordinary Differential Equations and Partial Differential Equations. For example : dy/dx + xy² = x²           It is an ordinary differential equation.                            ∂u/∂t + (∂u/∂x)² = 4           It is a partial differential equation.               I have already discussed the theorems related to ordinary differential equations. The theorems and equations are all important to the solutions of ordinary differential equations.  The geometrical structure for ordinary  differential equations...

    As I have discussed previously, there are many methods and related theorems to solve a differential equation.         So now I will discuss all theorems and applications related to it.It may be longer but beneficial for solving problems of differential equations.   Theorem_1 : The differential equation           M(x,y) dx + N(x,y)dy= 0 is exact iff          ∂M/∂y= ∂N/∂x Proof :                 If the given equation is exact , we have d(u(x,y))= M(x,y) dx + N(x,y) dy. ...   (1) and we know that d(u(x,y)) = ∂u/∂x dx + ∂u/∂y dy             .......(2) Consequently by comparison of (1) and(2) ∂u/ ∂x= M(x,y) , ∂u/ ∂y = N(x,y)          .........(3) Moreover,        ∂M/∂y=∂²u/ ∂y∂x and ∂N/∂x=∂²u/ ∂x∂y and because of ∂²u/∂y∂x =∂²u/∂x∂y, we get    ...

Hello my friends,             I want discuss something about the famous equation called LAGRANGE EQUATION.  This equation is used to solve differential equations of higher degree like CLAIRAUT EQUATION.         Now ,                  The equation y= xg(p)+ f(p) is associated with the name of JOSEPH LOUIS LAGRANGE (1736_1813).             This is a generalised form of CLAIRAUT's EQUATION .   If we put g(p) =p, it is in CLAIRAUT's form.             Differentiating w.r.t x and putting dy/ dy/dx = y'= p, we have         p= g(p) + xg'(p)  dp/dx+f'(p) dp/dx   or [p-g(p)] dx/dp= xg'(p)+ f'(p)    or        dx/dp= g'(p)x/p-g(p) + f'(p)/p-g(p)..(1)   The equation (1) is a linear differential equation in x and dx/dp and it is integrable using ...

Clairaut's Differential Equations :           The equation y= px+ f(p) .  .    .  .   . (1) of first degree in x and y is called the ' Claitaut's equation ' after the name of Alex Claude CLAIRAUT(1713_1765).        Differentiating w.r.t x, bearing in mind that p= dy/ dx, we obtain          p= xdp/dx+ p+ f(p)dp/ dx or     [x+f(p)]dp/dx=0. Equating each factor to zero, we get     dp/ dx=0.      .....(2) and x+ f(p)=0.     .......(3) Integrating (2) we obtain p=c ( a constant). Putting this value of p into (1), we find its complete integral    y= cx+ f(c).       ............(4)    which represents geometrically, a family of straight lines.   Elimination of p between (3) and (1) leads to a singular solution.   Example :           ...

Exact Differential Equations :             The differential equation of the form M(x,y)dx+ N(x,y)dy=0 is called exact differential equation if δM/δy=δN/δx.         And the differential equations of the form M(x,y)dx+ N(x,y)dy=0 is called non exact differential equations if δM/δy≠δN/δx.       And it's solution is given by  ∫M(x,y)dx + ∫N(x,y)dy=c     y as.           Terms don't contain                                       x constant Let's discuss some examples related to this form, Example :      1. Solve the differential equation     ( ycosx+ siny+y)dx+(sinx+ x cosy+x)dy=0 if it is exact. Solution :             Here M= ycosx+siny+y and N=sinx+xcosy+x  Therefore δM/δy= cosx+cosy+1 and δN/δx=cosx+cos...

Homogeneous Differential Equation:             Now I want to different forms and types of differential equations .First of all we should focus on the types, that are Homogeneous Differential equations, Exact and non exact differential equations etc. The homogeneous equation must have same degree.   For example :  solve                x^2y dx - ( x^3+y^3) dy = 0 Solution :              The given equation can be written in the form          dy/ dx = x^2y/x^3+y^3 Putting y= vx, we have   v+xdv/dx= x^2.vx/x^3+v^3x^3 = v/ 1+v^3 => x dv/dx = v/1+v^3 - v                     = -v^4/1+v^3 =>(1+v ³/v⁴)dv = - dx/x =>(1/v⁴+1/v)dv = - dx/x Integrating,      -1/3v³+log(v) = - log(x)+log(c) => log(vx/c)=1/3v³ => vx/c = exp(1/3v³) =...

Solution Of Differential Equations :     From ancient Greek Mathematics equations are the short form of a long analytical problem .  A large problem can be solved in a easy way by converting it into a simple equation.  Differential equations are the evolved form of equations .   So here i want to discuss about some problems of Differential equations and it's methods of solving problems .         First of all I want to say something about its definition.Its definition is that it is the equation containing or having differential coefficient .     For example_ :                   d^2y/dx^2+dy/dx+y=3     Let us consider some problems and solutions regarding this.   Examples :          1. d^2y/dx^2+y=x          2. y"+y'+6y=0 Solutions :           1. Given ...