Question

Show that the differential equation

2xz + q^2 = x(xp + yq)

has a complete integral


z+ a^2x = axy + bx^2

and deduce that

x(y + hx)^2 = 4(z - kx^2)

is also a complete integral.

Answer 1

Let $F(x,y,z,p,q)=2xz+q^2-x(xp+yq)=0.$Then Charpit’s equations are

$$\dfrac{dx}{-x^2 }=\dfrac{dy}{2q-xy}=\dfrac{dz}{-px^2+2q^2-qxy}=$$

$$=\dfrac{dp}{-2z+yq}=\dfrac{dq}{-qx}$$

From the first and last equations, we get

$$\dfrac{dx}{x }=\dfrac{dq}{q }$$

Integrate

$$\ln|x|+\ln a=\ln|q|$$

$$q=ax, a=const.$$

Then the equation $F(x, y, z, p, q)=0$ gives

$$2xz+a^2x^2-x(xp+ayx)=0$$

$$p=\dfrac{2z+a^2x-ayx}{x}$$

Hence the relation $dz=pdx+qdy$ leads to

$$dz=\dfrac{2z+a^2x-ayx}{x}dx+axdy=>$$

$$=>d(\dfrac{z}{x^2})+d(\dfrac{a^2}{x})=ad(\dfrac{y}{x})$$

which, on integration, gives

$$\dfrac{z}{x^2}+\dfrac{a^2}{x}=a\dfrac{y}{x}+b$$

$$z+a^2x=axy+bx^2$$

which is a complete integral, $b$ being a constant.

Now we show that

$$x(y + cx)^2 = 4(z − dx^2)\ \ \ \ \ \ \ \ \ (1)$$

is also a complete integral.

Let us consider the curve Γ : $y=0, z=\dfrac{c^2x^3+4dx^2}{4}$ on the surface $(1).$

At the intersections of $z+a^2x=axy+bx^2$ and the curve Γ we have

$$c^2x^2 + 4(d − b)x + 4a^2 = 0$$

which has equal roots when $b=d\pm ac.$ Taking $b=d+ac,$ the subsystem has the equation $z+a^2x=axy+(d+ac)x^2,$ i.e.

$$a^2x-x(cx+y)a+(z-dx^2)=0$$

which has the envelope

$$x^2(cx+y)^2=4x(z-dx^2),$$

i.e.

$$x(cx+y)^2=4(z-dx^2)$$

This is, therefore, a complete integral.