Answer to Question #256238 in Differential Equations for Sahil

Question

Pxy+Pq+qy=yz

Answer 1

Solve by Charpit's method

$$f(x,y,z,p,q)=pxy+pq+qy-yz=0$$

Then

$$f_x=py, f_y=px+q-z, f_z=-y,$$

$$f_p=xy+q, f_q=p+y$$

Charpit's auxiliary equations are

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

$$=\dfrac{dx}{-f_p}=\dfrac{dy}{-f_q}=\dfrac{df}{0}$$

$$\dfrac{dp}{py-py}=\dfrac{dq}{px+q-z-qy}=\dfrac{dz}{-(pf_p+qf_q)}$$

$$=\dfrac{dx}{-xy-q}=\dfrac{dy}{-p-y}$$

From first and second members

$$\dfrac{dp}{0}=\dfrac{dq}{px+q-z-qy}=>dp=0=>p=a$$

Then

$$pxy+pq+qy-yz=0$$

$$=>axy+aq+qy-yz=0$$

$$=>q=\dfrac{y(z-ax)}{a+y}$$

Now consider

$$dz=pdx+qdy=>dz=adx+\dfrac{y(z-ax)}{a+y}dy$$

Put $t=z-ax, dt=dz-adx$

$$dt=\dfrac{yt}{a+y}dy$$

$$\dfrac{dt}{t}=\dfrac{y}{a+y}dy$$

Integrate both sides

$$\int\dfrac{dt}{t}=\int\dfrac{y}{a+y}dy$$

$$\ln t=y-a\ln(a+y)+\ln C$$

The required solution is

$$\ln (z-ax)=y-a\ln(a+y)+\ln C$$