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.
hints on solving DE
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0Sorry for the $\frac{a}{2}$, it should be only a – 2012-08-02
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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0no ... it isn't given ... must be some arbitrary constant. thank you for your effort – 2012-08-02
Let
\begin{equation} {\frac{dx}{x}} = {\frac{dy}{y}} = {\frac{dz}{z - a \sqrt{x^2+y^2+z^2}}} = K \end{equation}
\begin{equation} {\frac{2xdx}{2x^{2}}} = {\frac{2ydy}{y^{2}}} = {\frac{2zdz}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}}} = K \end{equation}
Then \begin{equation} \frac{dx^{2}}{2x^{2}} = \frac{dy^{2}}{2y^{2}} = \frac{dz^{2}}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}} = K \end{equation}
Adding all the three terms, we get \begin{equation} \frac{dx^{2} + dy^{2} + dz^{2}}{2x^{2} + 2y^{2} + 2z^{2} - 2az \sqrt{x^2+y^2+z^2}} = K \end{equation}
\begin{equation} \frac{dw^{2}}{2w^{2} - 2az \sqrt{w^{2}}} = {\frac{2zdz}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}}} \end{equation}
Hence, \begin{equation} \frac{dw^{2}}{2w^{2} - 2az \sqrt{w^{2}}} = {\frac{dz}{z - a \sqrt{w^2}}} \end{equation}
And then, \begin{equation} \frac{2w dw}{2w^{2} - 2az w} = {\frac{dz}{z - a w}} \end{equation}
And so, \begin{equation} \frac{dw}{dz}= {\frac{w - az}{z - a w}} \end{equation}
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0Yeah, can happen. ;-) – 2012-08-02