Solving equation
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D
Deleted member 4993
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What do you want to do?
- solve for 'x' for a given 'y'
- solve for 'y' for a given 'x'
Please elaborate.
Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:
Please share your work/thoughts about this assignment.
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Deleted member 4993
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To plot the given expression, first make two-column table. Calculate 'x' for given different 'y's (what would be the minimum value of 'y'?).
Then plot.
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Deleted member 4993
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Is this an assignment from a class? Can you post the question EXACTLY (may be a picture of the assignment)?
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Deleted member 4993
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We wanted see the picture of the whole "problem statement".
I can't post any picture for copy write issue. Thanks in advance
HallsofIvy
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Subhotosh Khan's point is that, because the equation has x both in an exponential and not in an exponential, there is no solution in terms of elementary functions. There might be one in terms of "Lambert's W function", which is defined as the inverse function to \(\displaystyle f(x)= xe^x\). Do you know what that is?
And, while I am here, posting one problem from a textbook will NOT violate copyright laws. That would be "fair use"
And, while I am here, posting one problem from a textbook will NOT violate copyright laws. That would be "fair use"
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Sorry I don't know, that's why I asked for help. Can you please post a detailed solution
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Deleted member 4993
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Sorry,
We donot provide detailed solution - unless the student provides detailed "work" done!
HallsofIvy
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Sigh. You've been given a lot of help but don't yet know enough math to understand what you have been told. You have, for example, been told that there is no solution in terms of "elementary functions". Do you know what that means? Where did you get this equation? Do you understand that "almost all" algebraic equations cannot be solved in terms of elementary functions? I will "post a detailed solution" but it probably won't help you!
You are trying to solve the equation \(\displaystyle y^2- 2ln(y)= x^2\). That is the same as \(\displaystyle y^2- ln(y^2)=x^2\). Let \(\displaystyle u=y^2\) so that \(\displaystyle u- ln(u)= x^2\). Take the exponential of both sides- \(\displaystyle e^{u- ln(u)}=\)\(\displaystyle \frac{e^u}{u}=\)\(\displaystyle \frac{1}{ue^{-u}}= e^{x^2}\). Invert both sides. \(\displaystyle ue^{-u}= e^{-x^2}\). Finally, let v= -u. Then \(\displaystyle -ve^v= e^{-x^2}\) so \(\displaystyle ve^v= -e^{-x^2}\).
Now, as I said before, the "Lambert W function" is the inverse function to \(\displaystyle f(x)= xe^x\).
Applying the W function to both side of \(\displaystyle ve^{-v}= -e^{-x^2}\) we get \(\displaystyle v= W(-e^{-x^2})\). Then \(\displaystyle u= -v= -W(-e^{-x^2})\) and, finally, \(\displaystyle y= \pm\sqrt{-W(-e^{-x^2})}\).
\(\displaystyle y= \pm\sqrt{-W(-e^{-x^2})}\). Okay?
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Dr.Peterson
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If the goal is to graph the equation, why not do what can be done, and solve for x in terms of y? That makes it just as easy to plot as if you could solve for y. This is what you were told in #4, but you may not have noticed the point.
D
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That too for non-commercial use. That is generally a cop-out. When you have text-book, copying the problem for educational purposes cannot be construed as violation of copy-right law!!