# hints on solving DE

How to solve this DE? $${dx \over x} = {dy \over y} = {dz \over z - a \sqrt{x^2+y^2+z^2}}$$ From the first part, I get $y = c_1x$. How to find the other solution? The answer according to answer sheet is $z + \sqrt{x^2 + y^2 + z^2} = c_2$. Thank you for help.

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Try putting $w^{2} = x^{2} + y^{2} + z^{2}$ and then use the identity $$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{k_{1} a + k_{2} c + k_{3} e}{ k_{1} b+ k_{2} d+ k_{3} f}.$$ –  Jayesh Badwaik Aug 2 '12 at 11:39
@BabakSorouh the topic is "Simulatneous Equation of first order first degree" –  Santosh Linkha Aug 2 '12 at 11:41
@JayeshBadwaik wouldn't i get more variables from substitution? –  Santosh Linkha Aug 2 '12 at 11:44
No, basically you then eliminate $x$ and $y$ and solve for $w$ in terms of $z$. If you want more hint, I can give you so. –  Jayesh Badwaik Aug 2 '12 at 11:46
okay thanks ... let me try first. –  Santosh Linkha Aug 2 '12 at 11:48

## 2 Answers

$${dx \over x} = {dy \over y} = {dz \over z - a \sqrt{x^2+y^2+z^2}}$$ You get $y=c_1x$, so put it into the third fraction: $${dx \over x} = {dz \over z - a \sqrt{x^2+c_1^2x^2+z^2}}$$ $${dx \over x} = {dz \over z - a \sqrt{(1+c_1^2)x^2+z^2}}={dz \over z - a \sqrt{Cx^2+z^2}}$$ which is homogeneous equation: $$(z - a \sqrt{Cx^2+z^2})dx=xdz, x\neq 0$$ by taking $u=\frac{z}{x}$, you get: $${-adx \over x} = {du \over \sqrt{C+u^2}}$$ then integrating from both sides gives: $$\ln|u+\sqrt{C+u^2}|=-a\ln|x|+c_2$$ or $$\ln|z+\sqrt{x^2+y^2+z^2}|=(1-a)\ln|x|+c_2$$ Are you sure, you don't have any information about that $a$?

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no ... it isn't given ... must be some arbitrary constant. thank you for your effort –  Santosh Linkha Aug 2 '12 at 12:13

Let

$${\frac{dx}{x}} = {\frac{dy}{y}} = {\frac{dz}{z - a \sqrt{x^2+y^2+z^2}}} = K$$

$${\frac{2xdx}{2x^{2}}} = {\frac{2ydy}{y^{2}}} = {\frac{2zdz}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}}} = K$$

Then $$\frac{dx^{2}}{2x^{2}} = \frac{dy^{2}}{2y^{2}} = \frac{dz^{2}}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}} = K$$

Adding all the three terms, we get $$\frac{dx^{2} + dy^{2} + dz^{2}}{2x^{2} + 2y^{2} + 2z^{2} - 2az \sqrt{x^2+y^2+z^2}} = K$$

$$\frac{dw^{2}}{2w^{2} - 2az \sqrt{w^{2}}} = {\frac{2zdz}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}}}$$

Hence, $$\frac{dw^{2}}{2w^{2} - 2az \sqrt{w^{2}}} = {\frac{dz}{z - a \sqrt{w^2}}}$$

And then, $$\frac{2w dw}{2w^{2} - 2az w} = {\frac{dz}{z - a w}}$$

And so, $$\frac{dw}{dz}= {\frac{w - az}{z - a w}}$$

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looks like i really put up a different subs!! –  Santosh Linkha Aug 2 '12 at 12:27
Yeah, can happen. ;-) –  Jayesh Badwaik Aug 2 '12 at 12:40
thank both of you!! +2 –  Santosh Linkha Aug 2 '12 at 12:50