ksaimouli
  • ksaimouli
differentiate
Mathematics
schrodinger
  • schrodinger
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ksaimouli
  • ksaimouli
\[\frac{ dy }{ dx }=2x-y\]
anonymous
  • anonymous
@ksaimouli \[\frac{dy}{dx}+y=2x\]
ksaimouli
  • ksaimouli
i tried to do this |dw:1360374033568:dw|

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ksaimouli
  • ksaimouli
\[\int\limits_{}^{}dy+\int\limits_{}^{}y= \int\limits_{}^{}2x dx\]
ksaimouli
  • ksaimouli
u mean this
anonymous
  • anonymous
Can't do that, the y part has no dy in it, it won't make sense.
ksaimouli
  • ksaimouli
so how to do this
anonymous
  • anonymous
Well, start with\[\frac{dy}{dx}+y=2x\] and multiply everything by e^x
anonymous
  • anonymous
\[\large e^x\frac{dy}{dx}+e^xy=2xe^x\]
anonymous
  • anonymous
Now, question... what's \[\large \frac{d}{dx}ye^x\] ?
anonymous
  • anonymous
\[\large \frac{d}{dx}ye^x = e^x\frac{dy}{dx} + ye^{x}\]
ksaimouli
  • ksaimouli
i did not understant how did u get that ^
anonymous
  • anonymous
Implicit differentiation
ksaimouli
  • ksaimouli
|dw:1360374611401:dw|
ksaimouli
  • ksaimouli
hmm can u use implicit differentiation of function which is already differentiated
anonymous
  • anonymous
You can use it on any function, as far as I know :)
anonymous
  • anonymous
So, we end up with \[\large \frac{d(ye^x)}{dx}=2xe^x \]
anonymous
  • anonymous
so, just bring the dx on the other side... \[\large d(ye^x)=2xe^xdx\] And integrate both sides... \[\large ye^x = \int\limits_{}^{}2xe^x dx\] And you're good to go. :)
ksaimouli
  • ksaimouli
thx
agent0smith
  • agent0smith
@PeterPan I haven't done these in a while, but is it incorrect to just differentiate this to: \[\frac{ dy }{ dx }=2x-y\]\[\frac{ d^2y }{ dx^2 }=2-\frac{ dy }{ dx }\]\[\frac{ d^2y }{ dx^2 } + \frac{ dy }{ dx } =2\] All the original question said was: Differentiate:\[\frac{ dy }{ dx }=2x-y\]
anonymous
  • anonymous
I was wondering about that, but then again, ksaimouli put in "I tried to do this", with a drawing that shows an attempt to solve it as a differential equation, so.... yeah

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