Answer to Question #112040 in Differential Equations for Piyush vijay

Question

Solve charpit method: 2z+p^2+qy+2y^2=0

Answer 1

Let $f(x,y,z,p,q) \equiv 2z+p^2+qy+2y^2=0$ ............(1)

Charpit's auxiliary equation are

$$\frac{dp}{f_x+pf_z}=\frac{dq}{f_y+qf_z}=\frac{dz}{-pf_p-qf_q}=\frac{dx}{-f_p}=\frac{dy}{-f_q}$$

....................(2)

From (1), $f_x=0, \ f_y=q+4y, \ f_z=2, \ f_p=2p,\ f_q=y$ .......(3)

Using (3),(2) reduce to

$$\frac{dp}{2p}=\frac{dq}{3q+4y}=\frac{dz}{-2p^2-qy}=\frac{dx}{-2p}=\frac{dy}{-y}$$

Taking 1st and 4th fraction we get,

$$\implies \frac{dp}{2p}=\frac{dx}{-2p}$$

So that , $P=-x$ ..............(4)

Now taking 2nd and 5th fraction ,we get

$$\frac{dq}{3q+4y}=\frac{dy}{-y}$$

$$\implies \frac{dq}{dy}=-\frac{3q+4y}{y}$$

$$\implies \frac{dq}{dy}+\frac{3}{y}q=-4$$

Which is a linear equation in first order.

$$\therefore I.F=e^{\int \frac{3}{y} dy } =e^{3logy}=e^{log {y^3} }=y^3$$

Now , $q \ y^3=\int (-4)y^3dy =-y^4$

$\implies q=-y$ .

Now putting the value of p and q in

$dz=pdx+qdy$ we get $dz=-xdx-ydy$

$\implies 2z=-x^2-y^2+c$

Where "c" is a integration constant.