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sjerman1

  • 3 years ago

differential equations -> xe^-t dx/dt=t with the initial condition that x(0) = 1.

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  1. sjerman1
    • 3 years ago
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    \[xe ^{-t} \frac{ dx }{ dt }=t\]

  2. sjerman1
    • 3 years ago
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    \[\int\limits x dx=\int\limits \frac{ t }{ e^{-t}} dt \]

  3. sjerman1
    • 3 years ago
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    is where im at now, wondering if this is correct

  4. zepdrix
    • 3 years ago
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    \[\large \int\limits\limits x dx=\int\limits\limits t e^t\; dt\] Yup looks good so far. By Parts on the right side :D

  5. sjerman1
    • 3 years ago
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    \[\frac{ x^2 }{ 2 }+c=e^t(t-1)\]

  6. sjerman1
    • 3 years ago
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    right o got that now solving for c hold on

  7. sjerman1
    • 3 years ago
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    \[\frac{ 1 }{ 2 }+c=e^0(0-1)\] c=\[-\frac{ 3 }{ 2 }\]

  8. zepdrix
    • 3 years ago
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    yay good job c: Pshhhh you didn't need any help with this one lol

  9. sjerman1
    • 3 years ago
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    so now i have \[\frac{ x^2 }{ 2 }-\frac{ 3 }{ 2 }=e^t(t-1)\]

  10. sjerman1
    • 3 years ago
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    now the hard part for me is solving for x because the answer wants it in x= soooo what can i do to the 3/2? would that be added to the -1? so the right side would be e^t(t+1/2)?

  11. sjerman1
    • 3 years ago
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    \[\frac{ x^2 }{ 2 }=e^t(t-1)+\frac{ 3 }{ 2 }\]

  12. zepdrix
    • 3 years ago
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    Multiply both sides by 2\[\large x^2-3=2e^t(t-1)\]Add 3 to both sides,\[\large x^2=2e^t(t-1)+3\]Square root,\[\large x=\pm \sqrt{2e^t(t-1)+3}\]

  13. zepdrix
    • 3 years ago
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    That's probably the route I would take. Get rid of the ugly fraction first :D

  14. sjerman1
    • 3 years ago
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    ohhhh my that is what i was going to ask next. but multiplying by two is probaly the rout my prof would have taken. Thank you @zepdrix !

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