Answer to Question #146753 in Differential Equations for Nikhil

Question

Apply Charpit's method to solve the following differential equation:
p(1+q^2)=q(z-a)

Answer 1

$$Solution$$

Here, given equation is:

$$p(1+q^2)=q(z-a)\\ \therefore p+pq^2)=qz-qa\\ \implies p+pq^2-qz+qa=0---(1)$$

Therefore the Charpit's auxiliary equation is

$$\frac{dp}{\frac{df}{dx}+p\frac{df}{dz}}=\frac{dq}{\frac{df}{dy}+q\frac{df}{dz}}=\frac{dz}{-p\frac{df}{dp}-q\frac{df}{dq}}=\frac{dx}{-\frac{df}{dp}}=\frac{dy}{-\frac{df}{dq}}=\frac{dF}{0}$$

or

$$\frac{dp}{0+p(-q)}=\frac{dq}{0+q(-q)}=\frac{dz}{-p-pq^2}=\frac{dx}{-1-q^2}=\frac{dy}{-2pq-z}=\frac{dF}{0}$$

or

$$\frac{dp}{-pq}=\frac{dq}{-q^2}=\frac{dz}{-p-pq^2}=\frac{dx}{-1-q^2}=\frac{dy}{-2pq-z}=\frac{dF}{0}$$

So that

$$\implies \frac{dp-dq}{-pq+q^2}=\frac{dy-dx}{-2pq-z+1+q^2}\\ \implies \frac{dp-dq}{-pq+q^2}=\frac{dy-dx}{-pq+q^2}\\ \implies dp-dq = dy-dx\\ \therefore dp+dx=dy+dq\\$$

Integrating, we get

$$\implies p+x=y+q\\ \implies p=y-x+q\\$$

From equation (1), we get

$$p+pq^2-qz+qa=0 \implies p=pq^2-qz+qa\\ q(pq-z-a)=p\\ q=p\ or\\ q=\frac{z-a}{p}$$

Using the value $p$ and $q$ in $dz=pdx+qdy$

$$\implies dz=(y-x+q)dx+(\frac{z-a}{p})dy\\ \implies \frac{dz}{z}=(y-x+q)dx+(\frac{z-a}{p})dy$$

Integrating, we get

$$log\ z=yx+qx-\frac{x^2}{2}+\frac{y-ya}{py}$$