The Decision of three methods of the solutions $dx/P=dy/Q=dz/R$

Question

Question:

(A) $$\frac{adx}{(b-c)yz}= \frac{bdy}{(c-a)xz}=\frac{cdz}{(a-b)xy}$$

(B) $$\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{1-z^2}$$

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These are simultaneous diff eq. of the first order and the first degree in three variables

And I know that three methods of solution of $dx/P=dy/Q=dz/R$

First method:

Second method:

Third method:

I know them. But I cannot decide which method I need to use for above two questionss. I dont want to solve these two questions. Only I want to learn which methods I use respectively? Please give me suggestion or hint? Thank a lot.

Answer 1

The fraction notation you are using is a short hand, $$ \frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R} \Leftrightarrow \frac{dy}{dx} = \frac{Q}{P}; \frac{dz}{dx} = \frac{R}{P}; \text{ etc.}$$

Looking at your system from part (A) we can write,

$$ \frac{dy}{dx} = \frac{a}{(b-a)yz} \cdot \frac{(c-a)xz}{b} = \frac{a(c-a)}{b(b-a)} \frac{x}{y} = k \frac{x}{y}$$

This tells us that $$ \frac{1}{2}y^2 = \frac{k}{2} x^2 + C $$

We know nothing of $a,b,$ and $c$ so we should keep in mind that $k$ could be positive or negative. Keep in mind that we haven't found any curves yet this is a family of surfaces in three space, lets call it $S_1$. The integral curves are on the surfaces.

To find the integral curves we need another surface which we now they are on. We find this by relating another pair of fractions. Either the first with the last or the second with the last. The solution of this other DE will yield a different family of surfaces $S_2$.

The integral curves are then the curves defined by the intersection of surfaces from S_1 with surfaces from S_2.

If you need a parameterized from of the curves you can use the following approach.

Let f(x,y,z)=0 be a surface from $S_1$. Let $g(x,y,z)=0$ be a surface from $S_2$. Suppose these two surfaces have a nonzero intersection and that we want to find the curve which defines that intersection.

The normals of these surfaces are given by the gradients of $f$ and $g$. Furthermore the tangents of the integral curve are normal to these gradients since the curves travels along the surfaces. Therefore if the integral curve is given by $\vec{r}(t)$ then we can say that,

$$\frac{d}{dt} \vec{r}(t) = \nabla f \times \nabla g$$

This defines a autonomous system of differential equaionts relating the derivatives of $x,y,$ and $z with respect to time to themselves. I.e.,

$$ \frac{dx}{dt}= h(x,y,z) ; \text{ etc.} $$

For more on this read: "Partial Differential Equations with Applications" by Zachmanoglou and Dale W. Thoe

Answer 2

A.

$$\frac{a dx}{(b -c ) yz} = \frac{b dy}{(c -a ) xz} = \frac{c dx}{(a -b ) xy}$$

Now multiply the numerator and denominator first fraction by $x$ for second fraction by $y$ and third fraction by $z$ and add all of them and get it

$$\frac{ax dx}{(b-c)xyz} = \frac{by dy}{(c-a)xz} = \frac{cz dz}{(a-b)xy} = \frac{ax dx + by dy + cz dz}{(b-c+c-a + a - b) xyz} $$

So you shall get $ax dx + by dy + cz dz = 0$ from the last fraction. One solution $a\frac{x^2}{2} + b\frac{y^2}{2} + c\frac{z^2}{2} =\text{constant}$ will comes from here.

For other solution take any two of the fractions such as $$\frac{adx}{(b-c)yz} = \frac{bdy}{(c - a)xz}$$

and see one term will be cancelled from the denominator. So you shall get

$a(c-a)x dx = b(b-c)y dy$

i.e. $a(c-a)\frac{x^2}{2} - b(b-c)\frac{y^2}{2}$ = constant.

I hope one shall get it easily.

B.

$$\frac{dx}{xz - y} = \frac{dy}{yz - x} = \frac{dz}{1 - z^2}$$

Do it as follows.

$$\frac{dz}{1 - z^2} = \frac{dx + dy + 0 dz}{(x + y)z - (x+y) + 0. (1 - z^2)}$$

Simplify it and get one solution from the relation $\frac{dz}{1 - z^2} = \frac{d(x+y)}{x+y}$.

For he other solution take $dx - dy + 0dz$ instead of $dx + dy + 0 dz$.

I hope everybody will be satisfied.