(y^2+z^2-x^2) p-2 x y q=-2 z x

Name: Problem 13, Chapter 8: Differential Equations
Uploaded: 2022-01-12T06:37:58-08:00
Duration: 10 min 53 s
Channel: Wasim Sher
Description: (y^2+z^2-x^2) p-2 x y q=-2 z x

Key Concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are equations that involve unknown multivariable functions and their partial derivatives. They arise in various fields such as physics, engineering, and finance. The study of PDEs involves understanding how the derivatives relate to the original function, often to model phenomena like heat conduction, fluid flow, and wave propagation.

First Order Quasilinear Equations

First order quasilinear PDEs are a subclass of partial differential equations where the highest order derivatives appear linearly. Although the coefficients in front of these derivatives may depend on the independent variables and the unknown function, the linearity in the highest order terms allows methods such as the method of characteristics to simplify and solve the equation.

Method of Characteristics

The method of characteristics is a powerful technique used to solve first order PDEs by reducing them to a system of ordinary differential equations (ODEs) along characteristic curves. These curves are paths in the domain along which the PDE becomes an ODE, which can then be solved using standard methods. This approach transforms the problem into integrating along paths dictated by the coefficients of the differential operators.

Characteristic Curves

Characteristic curves are trajectories in the space of independent variables along which the behavior of the solution to a first order PDE is determined. By following these curves, one can track how information propagates in the domain and thereby reduce the PDE to a simpler form. These curves play a central role in both existence and uniqueness theories for PDEs and are key to applying the method of characteristics effectively.

Transcript

00:01 Dear students, here we have a partial differential equation which is y square plus z square minus x square of p minus 2xy of q is equal to minus 2x z.

00:14 So let's solve it.

00:17 So we know that our lagrange's auxiliary equations are so our lagrangus auxiliary equations are.

00:39 So we can write our lagrange's auxiliary equation as dx upon y square plus z square minus x square equals to dy upon minus of 2 x y and equals to dz upon minus of 2 x x z.

01:07 So this is our lagrangus auxiliary equations so we can write this as our first equation so now we'll take our second and third because we can simply solve these by just we can cancel these two terms and get our solution very easily so let's take take the last parts parts of auxiliary equation so which is dy upon minus two x y equals d z upon minus 2x z so we can simply cancel these two and we'll get d y upon y equals d z upon z so we can simply take the integration of d y upon y minus d z upon z equals to zero so our integration will be we know that one y by taking the integration of one over y it will give us log y and for one of z it will give us log z equal to c1 let's a constant be c1 so we'll write it as in the form of log c1 so we can write log terms as y upon z with equals to log c1 if you cancel these logs or c1 will be y upon or c1 will be y upon c so this is our c1 now we want to find c2 for c2 we will use our multiplier technique to find c2 for this using multipliers x y and z we get we'll get x d x plus y die y plus z t z upon x y try it probably it will give us by multiplying x we will get x y square plus x z square minus x cube minus 2 x y square minus 2 x z square so then by further simplifying it we will get x d x plus y die plus z plus z upon if you take x common so we get x square plus y square plus z square these two will get us x z square minus x z square and this two will give us minus y z square and if you take x common from this you get x square plus y square plus z square so then we can give this or the name of our second equation now we'll take our last two terms from equation 1...

Please give Ace some feedback