Question

Solve by Charpit's method : px+qy=z√1+pq​

Answer 1

Given: $px+qy=z\sqrt{1+pq}$

To find:

• solution by Charpit's method

Solution:

Here the given equation is

$F(x,y,z,p,q)=px+qy-z\sqrt{1+pq}$ .....(1)

∴ Charpit's auxiliary equations are

$\dfrac{dp}{\frac{\partial F}{\partial x}+p\frac{\partial F}{\partial z}}=\dfrac{dq}{\frac{\partial F}{\partial y}+q\frac{\partial F}{\partial z}}=\dfrac{dz}{-p\frac{\partial F}{\partial p}-q\frac{\partial F}{\partial q}}\\=\dfrac{dx}{-\frac{\partial F}{\partial p}}=\dfrac{dy}{-\frac{\partial F}{\partial q}}$

We take the first two ratios only:

$\quad \dfrac{dp}{\frac{\partial F}{\partial x}+p\frac{\partial F}{\partial z}}=\dfrac{dq}{\frac{\partial F}{\partial y}+q\frac{\partial F}{\partial z}}$

$\Rightarrow \dfrac{dp}{p+p(-\sqrt{1+pq})}=\dfrac{dq}{q+q(-\sqrt{1+pq})}$

$\Rightarrow \dfrac{dp}{p(1-\sqrt{1+pq})}=\dfrac{dq}{q(1-\sqrt{1+pq})}$

$\Rightarrow \dfrac{dp}{p}=\dfrac{dq}{q}$, where $p,q\neq 0$

Integrating, we get

$\quad \int \dfrac{dp}{p}=\int \dfrac{dq}{q}$

$\Rightarrow logp=logq+logc$, where $c$ is integral constant

$\Rightarrow logp=log(cq)$

$\Rightarrow p=cq$ .....(2)

Substituting $p=cq$ in (1), we get

$\quad cqx+qy=z\sqrt{1+cq^{2}}$

Squaring both sides, we get

$\quad c^{2}q^{2}+2cxyq^{2}+y^{2}q^{2}=z^{2}(1+cq^{2})$

$\Rightarrow (c^{2}+2cxy+y^{2}-cz^{2})q^{2}=z^{2}$

$\Rightarrow \{(cx+y)^{2}-cz^{2}\}q^{2}=z^{2}$

$\Rightarrow q^{2}=\dfrac{z^{2}}{(cx+y)^{2}-cz^{2}}$

$\Rightarrow \boxed{q=\dfrac{z}{\sqrt{(cx+y)^{2}-cz^{2}}}}$

Putting the value of $q$ in (2), we get

$\quad \boxed{p=\dfrac{cz}{\sqrt{(cx+y)^{2}-cz^{2}}}}$

Putting these values of $p$ and $q$ in

$\quad dz=pdx+qdy$, we get

$\Rightarrow dz=\dfrac{cz}{\sqrt{(cx+y)^{2}-cz^{2}}}dx\\+\dfrac{z}{\sqrt{(cx+y)^{2}-cz^{2}}}dy$

$\Rightarrow dz=\dfrac{z}{\sqrt{(cx+y)^{2}-cz^{2}}}(cdx+dy)$

$\Rightarrow \dfrac{dz}{z}=\dfrac{cdx+dy}{\sqrt{(cx+y)^{2}-cz^{2}}}$

Let $cx+y=\sqrt{c}t$ so that

$\quad cdx+dy=\sqrt{c}dt$

So we have

$\quad \dfrac{dz}{z}=\dfrac{\sqrt{c}dt}{\sqrt{ct^{2}-cz^{2}}}$

$\Rightarrow \dfrac{dz}{z}=\dfrac{\sqrt{c}dt}{\sqrt{c}\sqrt{t^{2}-z^{2}}}$

$\Rightarrow \dfrac{dz}{z}=\dfrac{dt}{\sqrt{t^{2}-z^{2}}}$

$\Rightarrow \dfrac{dt}{dz}=\dfrac{\sqrt{t^{2}-z^{2}}}{z}$

$\Rightarrow \dfrac{dt}{dz}-\dfrac{\sqrt{t^{2}-z^{2}}}{z}=0$

$\Rightarrow \dfrac{dt}{dz}-\sqrt{(\dfrac{t}{z})^{2}-1}=0$ .....(3)

This is a homogeneous equation.

We take: $t=vz$

Then $\dfrac{dt}{dz}=v+z\dfrac{dv}{dz}$

Continuing (3), we write

$\quad v+z\dfrac{dv}{dz}-\sqrt{v^{2}-1}=0$

$\Rightarrow z\dfrac{dv}{dz}=\sqrt{v^{2}-1}-v$

$\Rightarrow \dfrac{dz}{z}=\dfrac{dv}{\sqrt{v^{2}-1}-v}$

We multiply the numerator and the denominator of the right hand side of the above equation by the conjugate of $\sqrt{v^{2}-1}-v$ i.e. by $\sqrt{v^{2}-1}+v$ in order to rationalize the denominator:

$\quad \dfrac{dz}{z}=\dfrac{\sqrt{v^{2}-1}+v}{(\sqrt{v^{2}-1}-v)(\sqrt{v^{2}-1}+v)}dv$

$\Rightarrow \dfrac{dz}{z}=\dfrac{\sqrt{v^{2}-1}+v}{v^{2}-1-v^{2}}dv$

$\Rightarrow \dfrac{dz}{z}=\dfrac{\sqrt{v^{2}-1}+v}{-1}dv$

$\Rightarrow \dfrac{dz}{z}=-\sqrt{v^{2}-1}dv-vdv$

Integrating, we get

$\quad \int \dfrac{dz}{z}=-\int \sqrt{v^{2}-1}dv-\int vdv$

$\Rightarrow logz=-[\dfrac{v(\sqrt{v^{2}-1})}{2}\\-\dfrac{1}{2}log(v+\sqrt{v^{2}-1})]-\dfrac{v^{2}}{2}+k$

where $k$ is constant of integration

$\Rightarrow logz=-\dfrac{cx+y}{2\sqrt{c}z}\sqrt{(\dfrac{cx+y}{\sqrt{c}z})^{2}-1}\\+\dfrac{1}{2}log[\dfrac{cx+y}{\sqrt{c}z}+\sqrt{(\dfrac{cx+y}{\sqrt{c}z})^{2}-1}]\\-\dfrac{1}{2}(\dfrac{cx+y}{\sqrt{c}z})^{2}+k$

where $v=\dfrac{cx+y}{\sqrt{c}z}$

This is the required solution.