## sjerman1 2 years ago differential equations -> xe^-t dx/dt=t with the initial condition that x(0) = 1.

1. sjerman1

$xe ^{-t} \frac{ dx }{ dt }=t$

2. sjerman1

$\int\limits x dx=\int\limits \frac{ t }{ e^{-t}} dt$

3. sjerman1

is where im at now, wondering if this is correct

4. zepdrix

$\large \int\limits\limits x dx=\int\limits\limits t e^t\; dt$ Yup looks good so far. By Parts on the right side :D

5. sjerman1

$\frac{ x^2 }{ 2 }+c=e^t(t-1)$

6. sjerman1

right o got that now solving for c hold on

7. sjerman1

$\frac{ 1 }{ 2 }+c=e^0(0-1)$ c=$-\frac{ 3 }{ 2 }$

8. zepdrix

yay good job c: Pshhhh you didn't need any help with this one lol

9. sjerman1

so now i have $\frac{ x^2 }{ 2 }-\frac{ 3 }{ 2 }=e^t(t-1)$

10. sjerman1

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

$\frac{ x^2 }{ 2 }=e^t(t-1)+\frac{ 3 }{ 2 }$

12. zepdrix

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

That's probably the route I would take. Get rid of the ugly fraction first :D

14. sjerman1

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 !