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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[xe ^{t} \frac{ dx }{ dt }=t\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits x dx=\int\limits \frac{ t }{ e^{t}} dt \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is where im at now, wondering if this is correct

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large \int\limits\limits x dx=\int\limits\limits t e^t\; dt\] Yup looks good so far. By Parts on the right side :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ x^2 }{ 2 }+c=e^t(t1)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right o got that now solving for c hold on

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 2 }+c=e^0(01)\] c=\[\frac{ 3 }{ 2 }\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1yay good job c: Pshhhh you didn't need any help with this one lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so now i have \[\frac{ x^2 }{ 2 }\frac{ 3 }{ 2 }=e^t(t1)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now 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)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ x^2 }{ 2 }=e^t(t1)+\frac{ 3 }{ 2 }\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Multiply both sides by 2\[\large x^23=2e^t(t1)\]Add 3 to both sides,\[\large x^2=2e^t(t1)+3\]Square root,\[\large x=\pm \sqrt{2e^t(t1)+3}\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1That's probably the route I would take. Get rid of the ugly fraction first :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohhhh 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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