Answer to Question #154962 in Differential Equations for ANIK KUMAR GHOSH

Question

xpq+yq^2=1 using charpits method

Answer 1

$xpq+yq^2=1\\ xpq+yq^2-1=0.......(1)\\$

Compare (1) to $f(x,y,z,p,q)=0$

$f(x,y,z,p,q)=xpq+yq^2-1=0.......(2)$

Now, the Charpit's auxiliary equation is:

$\frac{dx}{-\frac{\partial f}{\partial p}}=\frac{dy}{-\frac{\partial f}{\partial q}}=\frac{dz}{-p\frac{\partial f}{\partial p}-q\frac{\partial f}{\partial p}}=\frac{dp}{\frac{\partial f}{\partial x}+p\frac{\partial f}{\partial z}}=\frac{dq}{\frac{\partial f}{\partial y}+q\frac{\partial f}{\partial z}}\\ \frac{\partial f}{\partial x}=xq\\ \frac{\partial f}{\partial y}=q^2\\ \frac{\partial f}{\partial z}=0\\ \frac{\partial f}{\partial p}=xq\\ \frac{\partial f}{\partial q}=xp+2yq$

So, the Chapit's equation is;

$\frac{dp}{pq+p.0}=\frac{dq}{q^2+q.0}=\frac{dz}{-p.xq-q.(xp+2yq)}=\frac{dy}{-(xp+2yp)}=\frac{dx}{-xq}\\ \text{Take the second and fifth fractions}\\ \frac{dq}{q^2}=\frac{dx}{-xq}\\ \frac{dq}{q}+\frac{dx}{x}=0\\ \text{Integrate}\\ \int \frac{dq}{q}+\int \frac{dx}{x}=\int0\\ \ln q+\ln x=\ln a\\ q=\frac{a}{x}\\ \text{Substitute this into (2)}\\ ax^2p+ya^2-x^2=0\\ p=\frac{x^2-ya^2}{ax^2}\\ \text{Substitute the value of p and q into } dz=pdx+qdy\\ dz=\frac{(x^2-a^2y)}{ax^2}dx+\frac{a}{x}dy\\ dz=(\frac{1}{a}-\frac{ay}{x^2})dx+\frac{a}{x}dy\\ \text{Integrate}\\ z=\frac{x}{a}+\frac{ay}{x}+\frac{ay}{x}+b\\ z=\frac{x}{a}+\frac{2ay}{x}+b \text{ where a, b are arbitrary constants}$