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November 27th, 2017, 07:52 AM   #1
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Singular solution

How to show that $\displaystyle t=0$ is a solution of $\displaystyle 2t^2e^{-x}\,dt+x\sqrt{t}\,dx=0$? I do not know how to calculate $\displaystyle dt$ when $\displaystyle t=0$. Thanks.

Last edited by eulid; November 27th, 2017 at 08:18 AM.
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November 27th, 2017, 09:51 AM   #2
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The way you have phrased this is a little confusing. A first order equation like this can be thought of as an equation describing x as a function of t or as an equation describing t as a function of x. The best I can do in interpreting your question is that you want to show that t, as a function of x, is the constant function t= 0 no matter what x is. In that case, since t is a constant, the derivative is 0: dt/dx= 0 so dt= 0 while dx can be anything. Set t= 0 and dt= 0 and show that the equation is satisfied for all x.
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November 27th, 2017, 11:17 AM   #3
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You don't even need $\mathrm dt$ or $\frac {\mathrm dt}{\mathrm dx}$ for this. $t=0$ is sufficient.
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