Here's the question you clicked on:
ksaimouli
differentiate
\[\frac{ dy }{ dx }=2x-y\]
@ksaimouli \[\frac{dy}{dx}+y=2x\]
i tried to do this |dw:1360374033568:dw|
\[\int\limits_{}^{}dy+\int\limits_{}^{}y= \int\limits_{}^{}2x dx\]
Can't do that, the y part has no dy in it, it won't make sense.
Well, start with\[\frac{dy}{dx}+y=2x\] and multiply everything by e^x
\[\large e^x\frac{dy}{dx}+e^xy=2xe^x\]
Now, question... what's \[\large \frac{d}{dx}ye^x\] ?
\[\large \frac{d}{dx}ye^x = e^x\frac{dy}{dx} + ye^{x}\]
i did not understant how did u get that ^
Implicit differentiation
|dw:1360374611401:dw|
hmm can u use implicit differentiation of function which is already differentiated
You can use it on any function, as far as I know :)
So, we end up with \[\large \frac{d(ye^x)}{dx}=2xe^x \]
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. :)
@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:\[\frac{ dy }{ dx }=2x-y\]
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