Question

Form a Partial differential eq. of ᵩ(x+y+u, x2+y2+u2)=0

Answer 1

Answer on Question #68374 – Math – Differential Equations

Question

Form a partial differential equation of

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

Solution

The solution of partial differential equation satisfies

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

for some arbitrary function $\varphi$ and independent first integrals

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

Differentiating first integrals with respect to $x$

$$\left\{ \begin{array}{c} 1 + \frac {d y}{d x} + \frac {d u}{d x} = 0, \\ 2 x + 2 y \frac {d y}{d x} + 2 u \frac {d u}{d x} = 0, \end{array} \right.$$ $$\left\{ \begin{array}{l} \frac {d y}{d x} + \frac {d u}{d x} = - 1, \\ y \frac {d y}{d x} + u \frac {d u}{d x} = - x. \end{array} \right.$$

Using Cramer's rule

$$\frac {d y}{d x} = \frac {\left| \begin{array}{cc} - 1 & 1 \\ - x & u \end{array} \right|}{\left| \begin{array}{cc} 1 & 1 \\ y & u \end{array} \right|} = \frac {- u + x}{u - y},$$ $$\frac {d u}{d x} = \frac {\left| \begin{array}{cc} 1 & - 1 \\ y & - x \end{array} \right|}{\left| \begin{array}{cc} 1 & 1 \\ y & u \end{array} \right|} = \frac {- x + y}{u - y}.$$

Then

$$\frac {\frac {d y}{- u + x}}{u - y} = d x,$$ $$\frac {\frac {d u}{- x + y}}{u - y} = d x,$$

hence

$$dx = \frac{dy}{\frac{-u + x}{u - y}} = \frac{du}{\frac{-x + y}{u - y}}$$

that corresponds to the partial differential equation

$$\frac{\partial u}{\partial x} + \frac{-u + x}{u - y} \frac{\partial u}{\partial y} = \frac{-x + y}{u - y},$$ $$(u - y) \frac{\partial u}{\partial x} + (x - u) \frac{\partial u}{\partial y} = (y - x).$$

Answer: $(u - y) \frac{\partial u}{\partial x} + (x - u) \frac{\partial u}{\partial y} = (y - x)$.

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