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ksaimouli

  • 2 years ago

differentiate

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  1. ksaimouli
    • 2 years ago
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    \[\frac{ dy }{ dx }=2x-y\]

  2. PeterPan
    • 2 years ago
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    @ksaimouli \[\frac{dy}{dx}+y=2x\]

  3. ksaimouli
    • 2 years ago
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    i tried to do this |dw:1360374033568:dw|

  4. ksaimouli
    • 2 years ago
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    \[\int\limits_{}^{}dy+\int\limits_{}^{}y= \int\limits_{}^{}2x dx\]

  5. ksaimouli
    • 2 years ago
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    u mean this

  6. PeterPan
    • 2 years ago
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    Can't do that, the y part has no dy in it, it won't make sense.

  7. ksaimouli
    • 2 years ago
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    so how to do this

  8. PeterPan
    • 2 years ago
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    Well, start with\[\frac{dy}{dx}+y=2x\] and multiply everything by e^x

  9. PeterPan
    • 2 years ago
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    \[\large e^x\frac{dy}{dx}+e^xy=2xe^x\]

  10. PeterPan
    • 2 years ago
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    Now, question... what's \[\large \frac{d}{dx}ye^x\] ?

  11. PeterPan
    • 2 years ago
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    \[\large \frac{d}{dx}ye^x = e^x\frac{dy}{dx} + ye^{x}\]

  12. ksaimouli
    • 2 years ago
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    i did not understant how did u get that ^

  13. PeterPan
    • 2 years ago
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    Implicit differentiation

  14. ksaimouli
    • 2 years ago
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    |dw:1360374611401:dw|

  15. ksaimouli
    • 2 years ago
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    hmm can u use implicit differentiation of function which is already differentiated

  16. PeterPan
    • 2 years ago
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    You can use it on any function, as far as I know :)

  17. PeterPan
    • 2 years ago
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    So, we end up with \[\large \frac{d(ye^x)}{dx}=2xe^x \]

  18. PeterPan
    • 2 years ago
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    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. :)

  19. ksaimouli
    • 2 years ago
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    thx

  20. agent0smith
    • 2 years ago
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    @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:\[\frac{ dy }{ dx }=2x-y\]

  21. PeterPan
    • 2 years ago
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    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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