Answer to Question #289455 in Differential Equations for Pankaj

Question

Using Charpit’s method, solve:

P²+ q² -2px -2qy +1=0

Answer 1

Solution:

$f(x, y, z, p, q)=p^{2}+q^{2}-2 p x-2 q y+1=0$

Charpit's auxiliary equation are:

$\frac{d p}{f_{x}+p f_{z}}=\frac{d q}{f_{y}+q f_{z}}=\frac{d z}{-p f_{p}-q f_{q}}=\frac{d x}{-f_{p}}=\frac{d y}{-f_{q}}$

or

$\frac{d p}{p}=\frac{d q}{q}=\frac{d z}{p(p-x)+q(q-y)}=\frac{d x}{p-x}=\frac{d y}{q-y}$

Taking first two fractions,

$\frac{d p}{p}=\frac{d q}{q}$

Integrating, $\ln p=\ln q+\ln a \quad or \quad p=a q.$

Putting $p=a q$ into f(x, y, z, p, q), we get:

$\begin{aligned} &a^{2} q^{2}+q^{2}-2 a q x-2 q y+1=0 \\ &q^{2}\left(a^{2}+1\right)-2 q(a x+y)+1=0 \end{aligned}$

Solve for q

$q=\frac{2(a x+y) \pm \sqrt{4(a x+y)^{2}-4\left(a^{2}+1\right)}}{2\left(a^{2}+1\right)}$

Let's $(a x+y)=u$ and $\left(a^{2}+1\right)=b=$ const

$q=\frac{u \pm \sqrt{u^{2}-b}}{b}$

$p=a q=\frac{a}{b}\left(u \pm \sqrt{u^{2}-b}\right)$

We know that

$d z=p d x+q d y=a q d x+q d y=q(a d x+d y)=q d(a x+y)=q d u$

Integrating

$z=\int \frac{u \pm \sqrt{u^{2}-b}}{b} d u=\frac{u^{2}}{2 b} \pm \frac{1}{b} \int \sqrt{u^{2}-b} d u=$

$\begin{aligned} &\quad=\frac{u^{2}}{2 b} \pm \frac{1}{2} u \sqrt{u^{2}-b}-\frac{1}{2} b \ln \left(\sqrt{u^{2}-b}+u\right)+C= \\ &\quad=\frac{(a x+y)^{2}}{2 b} \pm \frac{1}{2}(a x+y) \sqrt{(a x+y)^{2}-b}-\frac{1}{2} b \ln \left(\sqrt{(a x+y)^{2}-b}+(a x+y)\right)+C \end{aligned}$