Question

Find the differential equation of the family of surfaces fi(x+y+z, x2+y2-z2)=0 .
What is the order of this p.d.e ?

Answer 1

Answer on Question #40487 – Math - Differential Calculus

We have

$$\varphi (x ^ {2} + y ^ {2} - z ^ {2}, x + y + z) = 0$$

or the equations

$$x ^ {2} + y ^ {2} - z ^ {2} = c _ {1},$$ $$x + y + z = c _ {2}.$$

By differentiating we get

$$x d x + y d y - z d z = 0,$$ $$d x + d y + d z = 0$$

or the equations of the form

$$x P + y Q - z R = 0,$$ $$P + Q + R = 0.$$

Our differential equation in the standard form

$$P p + Q q = R$$

or

$$\frac {d x}{P} = \frac {d y}{Q} = \frac {d z}{R}$$

Each fraction is equal to

$$\frac {P _ {1} d x + Q _ {1} d y + R _ {1} d z}{P _ {1} P + Q _ {1} Q + R _ {1} R}$$

Multipliers may be chosen such that the numerator $P_{1}dx + Q_{1}dy + R_{1}dz$ is an exact differential of the denominator $P_{1}P + Q_{1}Q + R_{1}R$.

Let

$$P = z + y, Q = x + z, R = y - x$$

Then

$$d \big (P _ {1} (z + y) + Q _ {1} (x + z) + R _ {1} (y - x) \big) = P _ {1} d x + Q _ {1} d y + R _ {1} d z$$

So we have the first-order differential equation

$$(z + y) p + (x + z) q = y - x.$$