Answer to Question #220156 in Differential Equations for Efiii jaan

Question

find the characteristics of the equation pq=xy and determine the integral surface which passes through curve z=x, y=0

Answer 1

Solving with method of characteristics :

$$f(x, y, z, p, q)=pq-xy=0$$

$$f_x=-y, f_y=-x, f_z=0, f_p=q, f_q=p$$

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

$$\dfrac{dx}{q}=\dfrac{dy}{p}=\dfrac{dz}{2pq}=\dfrac{dp}{y}=\dfrac{dq}{x}=dt$$

Solve

$$\dfrac{dp}{dt}=y, \dfrac{dy}{dt}=p$$

$$\dfrac{d^2p}{dt^2}=\dfrac{dy}{dt}$$

$$\dfrac{d^2p}{dt^2}=p$$

$$p=Ae^{t}+Be^{-t}$$

$$y=\dfrac{dp}{dt}=Ae^{t}-Be^{-t}$$

$$q=Ce^{t}+De^{-t}$$

$$x=\dfrac{dq}{dt}=Ce^{t}-De^{-t}$$

$$\dfrac{dz}{dt}=2pq=2(Ae^{t}+Be^{-t})(Ce^{t}+De^{-t})$$

$$=2ACe^{2t}+2BDe^{-2t}+2AD+2BC$$

$$z=ACe^{2t}-BDe^{-2t}+2ADt+2BCt+E$$

$$pq-xy=0=>(Ae^{t}+Be^{-t})(Ce^{t}+De^{-t})$$

$$-(Ae^{t}-Be^{-t})(Ce^{t}-De^{-t})=0$$

$$=>AD+BC=0$$

The initial data curve is written in parametric form as

$$x_0(s)=s, y_0(s)=0, z_0(s)=s$$

Thus, using the initial data, the given PDE becomes

$$p_0(s)q_0(s)=0$$

The strip condition gives

$$1=p_0(1)+q_0(0)$$

Therefore we have

$$x_0=s, y_0=0, z_0=s,p_0=1, q_0=0$$

Substitute

$$C-D=s$$

$$A-B=0$$

$$AC-BD+E=s$$

$$A+B=1$$

$$C+D=0$$

$$A=\dfrac{1}{2}, B=\dfrac{1}{2}, C=\dfrac{1}{2}s, D=-\dfrac{1}{2}s, E=\dfrac{1}{2}s$$

Characteristics:

$$x=\dfrac{e^t+e^{-t}}{2}s$$

$$y=\dfrac{e^t-e^{-t}}{2}$$

$$z=\dfrac{1}{2}(\dfrac{e^{2t}+e^{-2t}}{2})s+\dfrac{1}{2}$$

$$p=\dfrac{e^t+e^{-t}}{2}$$

$$q=\dfrac{e^t-e^{-t}}{2}s$$

$$x=s\text{cosh}t, y=\text{sinh}t$$

$$x^2=s^2\text{cosh}^2t , 1+y^2=1+\text{sinh}^2t=\text{cosh}^2t$$

$$z=\dfrac{1}{2}s(\text{cosh}(2t)+1)=s\text{cosh}^2t$$

$$z^2=s^2\text{cosh}^4t=x^2(1+y^2)$$

Integral Surface: 

$$z^2=x^2(1+y^2)$$