Answer to Question #218186 in Differential Equations for Efiii jaan

Question

Find the characteristics of the equation pq=xy
And determine the integral surface passes through the curve z=x , y=0

Answer 1

We can also solve this question by method of separation of variables . We look for particular solution on the form -

$z(x,y)=f(x)g(y)$

$=(g\dfrac{df}{dx})(f\dfrac{dg}{dy})=xy\implies$ $fdf=xdx$ $\implies$ $f={\pm}\sqrt{x^{2}+c_1}$

$gdg=ydy$ $\implies$ $g={\pm}\sqrt{y^{2}+c_2}$

So the function of z will become as -

$z(x,y)=$ ${\pm}$ $\sqrt{x^{2}+c_1}\sqrt{y^{2}+c_2}$

With condition z$(x,0)=x={\pm}\sqrt{(x^{2}+c_1)c_2}$ $\implies$ $c_1=0;c_2=1;$ sign of $x$

$z(x,y)=x\sqrt{y^{2}+1}$

Solving with method of characteristics we get -

$=\dfrac{dx}{q}=\dfrac{dy}{p}=\dfrac{dz}{2pq}=\dfrac{dp}{y}=\dfrac{dq}{x}=\dfrac{dt}{1}$

$=x_0=s,y_0=0,z_0=s,p_0=1,q_0=0$

Solving the system leads to -

$x=Ae^{t}+Be^{-t}$

$y=Ce^{t}+De^{-t}$

$p=Ce^{2t}-De^{2t}$

$q=Ae^{2t}-Be^{2t}$

$z=ACe^{2t}-BDe^{2t}+E$

$=pq-xy=0$ $\implies$ $AB+BC=0$

with boundary conditions ; $p_0=1; q_0=0$

A$+B=s;\ C+D=0$ $;AC-BD+E=s$ $;C-D=1;A-B=0;AD+BC=0$

$A=B=E=\dfrac{s}{2}$ and $C=-D=\dfrac{1}{2}\implies$ $x=hcosht$

$y=sin(ht)$

$z=\dfrac{s}{2}(cosh(2t)+1)=scosh^{2}t$

$x^{2}-y^{2}s^{2}=s\implies$ $s^{2}=\dfrac{x^{2}}{y^{2}+1}$

$z=\dfrac{x^{2}}{s}$ $=x^{2}(\dfrac{+\sqrt{y^{2}+1}}{x})$

$z=x\sqrt{y^{2}+1}$