Question

Given that z = ax +sqrt (a^2 − 4y) + c is the complete integral of the PDE, p^2 − q^2 =4
determine its general integral.

Answer 1

Correction.

To understand the recall some theory.

Equation of the form $f(p,q)=0$ i.e. the equation contains $p$ and $q$ only.

Suppose that $z(x,y)=ax+by+c$ is solution of equation, then

$$p=\frac{\partial z}{\partial x}=a\,\,\,\text{and}\,\,\,q=\frac{\partial z}{\partial y}=b$$

substitute the above in equation $f(p,q)=0$ , we get $f(a,b)=0$ on solving this we can get $b=\phi(a)$ , where $\phi$ is known function. Using $b=\phi(a)$ , the complete solution of the given partial equation is

$$z=ax+\phi(a)y+c$$

In our case,

$$p^2-q^2=4\longrightarrow\underbrace{ p^2-q^2-4}_{f(p,q)}=0$$

Then, suppose

$$z=ax+by+c\\[0.3cm] p=\frac{\partial z}{\partial x}=a\,\,\,\text{and}\,\,\,q=\frac{\partial z}{\partial y}=b\\[0.3cm] p^2-q^2-4=0\longrightarrow a^2-b^2-4=0\longrightarrow\\[0.3cm] \boxed{b=\pm\sqrt{a^2-4},\,\,\,\forall |a|\ge2}$$

Conclusion,

$$\boxed{z=ax\pm\sqrt{a^2-4}\cdot y+c\,\,\,-\,\,\text{is the general integral}}$$

and

$$z=ax+\sqrt{a^2-4}\cdot y+c$$

is the complete integral as a special case.