if we have $\displaystyle dy/dx=e^(2x-1)$ how do we find the general solution??? help much appreciated!! thankyou!!

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- Feb 13th 2010, 10:16 AMYehiaDifferential equations, separating variables!
if we have $\displaystyle dy/dx=e^(2x-1)$ how do we find the general solution??? help much appreciated!! thankyou!!

- Feb 13th 2010, 10:43 AMNyrox
It's a matter of simple integration. You know that $\displaystyle [e^x]'=e^x$, so

$\displaystyle [\frac{1}{2}e^{2x-1}]'=e^{2x-1}$

A general solution is thus

$\displaystyle y(x)=\frac{1}{2}e^{2x-1}+C$ (with $\displaystyle C\in \mathbb R$ - Feb 13th 2010, 10:58 AMYehia
Yes thank you!! :) that's right, can you just help me with 2 other problems?

firstly, if i have the function cosx-sinx=0 how do i solve it? im stumped :/

and secondly how to i find the volum of space if the function x^(1/2)*e^x revolved around the x axis 360 degrees with the limits 1 and 2?

Thanks a million!!!!!(Rofl) - Feb 13th 2010, 11:04 AMicemanfan
- Feb 13th 2010, 11:09 AMNyrox
These have nothing to do with your original question... But anyway, for the first one, plot the graphs of both sine and consine, and you'll get your answer. As for the second, you must've seen the connection between the integral and a solid obtain by revolution. Calling $\displaystyle V$ the volume of the solid obtained by revolving a curve $\displaystyle f(x)$ around the $\displaystyle x$ axis between points $\displaystyle a$ and $\displaystyle b$:

$\displaystyle V=\int_a^b\pi (f(x))^2 \, dx$ - Feb 13th 2010, 11:12 AMANDS!
- Feb 13th 2010, 11:17 AMYehia
- Feb 13th 2010, 11:42 AMANDS!
I see no method to solve that problem unfortunately. Integration by parts will get you no where, and there is no method (that I know of), that will let you itegrate that sucker easily. What level math is this from? I can not imagine, if you are doing volumes of solids of revolution - which is Integral Calculus (or Calc 2), that they would ask you to do this problem.