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

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

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: enter image description here

Second method: enter image description here

Third method: enter image description here

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.

  • $\begingroup$ Think a little. You do not require to remember so many methods. $\endgroup$
    – Supriyo
    Commented Oct 13, 2013 at 13:52
  • $\begingroup$ My instructor told only these there methods. I dont know others. I just have been learning these subject. I know how to apply these methods to the equations but I dont decide which methods I need to use. Please can you help me? Dear @Samprity $\endgroup$
    – 1190
    Commented Oct 13, 2013 at 13:53
  • $\begingroup$ Accourding to you, which one should I use to solve these equations respectively? @Samprity $\endgroup$
    – 1190
    Commented Oct 13, 2013 at 14:01
  • $\begingroup$ Hope it is clear now. $\endgroup$
    – Supriyo
    Commented Oct 13, 2013 at 14:06
  • $\begingroup$ $(\text{A})$ comes from the PDE $\dfrac{(b-c)yz}{a}\dfrac{\partial z}{\partial x}+\dfrac{(c-a)xz}{b}\dfrac{\partial z}{\partial y}=\dfrac{(a-b)xy}{c}$ and $(\text{B})$ comes from the PDE $(xz-y)\dfrac{\partial z}{\partial x}+(yz-x)\dfrac{\partial z}{\partial y}=1-z^2$ ? $\endgroup$ Commented Oct 13, 2013 at 16:44

2 Answers 2



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

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.


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

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.

  • $\begingroup$ That's for part-A, you use first method. Right? And for part-B, I dont understand which method you used:( sorry Dear @Samprity $\endgroup$
    – 1190
    Commented Oct 13, 2013 at 14:13
  • $\begingroup$ I did not see all of your methods. $\endgroup$
    – Supriyo
    Commented Oct 13, 2013 at 14:15
  • $\begingroup$ Hmm, but I also dont know your solution methods:( can you solve more clear? I am a new learner of PDE :( $\endgroup$
    – 1190
    Commented Oct 13, 2013 at 14:16
  • 1
    $\begingroup$ For solution of A method a, then method c, and for B, method b only. $\endgroup$
    – Supriyo
    Commented Oct 13, 2013 at 14:48
  • $\begingroup$ Okay thank you I am trying right now:) $\endgroup$
    – 1190
    Commented Oct 13, 2013 at 14:52

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

  • $\begingroup$ Okay thank you for helping. I read more carefully, and solve these.:) Dear @Spencer $\endgroup$
    – 1190
    Commented Oct 14, 2013 at 0:46
  • $\begingroup$ Well, I have some questions. Does each of these two questions have only exact one integral curve? Or do there exist various integral curve equations? $\endgroup$
    – 1190
    Commented Oct 14, 2013 at 0:54
  • $\begingroup$ Also, do you know a book which I'm able to understand this subject better? $\endgroup$
    – 1190
    Commented Oct 14, 2013 at 0:55
  • $\begingroup$ I explain my first question with an example. For example, for question-B, my book answer says $(x+y)(z+1)=c_1$ and $(x-y)(z-1)=c_2$. But I found that $x+yz=c_1$ is this possible? $\endgroup$
    – 1190
    Commented Oct 14, 2013 at 0:59
  • 1
    $\begingroup$ @B11b it was my pleasure. I'm glad I was able to help :). $\endgroup$
    – Spencer
    Commented Oct 14, 2013 at 17:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .