first order differential equations

Original post by Sina800

It's still not working out. The answer is s=( ¹/₂t² + k ) e^3t
This is my working out:
^d/_dt(-3s) = -3te^3t

Integration by parts for the right hand side:

U= -t du= -1 dv= 3e^3tv= e^3t

-te^3t- ∫ -1_*e^3t
=-te^3t+ ∫ e^3t
= -te^3t + ¹/₃e^3t+ K

-3s= ¹/₃e^3t- te^3t + K

the answer is correct!

you have not multiplied the right hand side by the integrating factor!

Reply 6

OP

Original post by TeeEm

the answer is correct!

you have not multiplied the right hand side by the integrating factor!

I have multiplied RHS if my integrating factor is u(t) = -3

^d/_dt(-3s) = -3te^3t

And I've followed the steps but I don't see where the t² has come from.

Reply 7

Original post by Sina800

I have multiplied RHS if my integrating factor is u(t) = -3

^d/_dt(-3s) = -3te^3t

And I've followed the steps but I don't see where the t² has come from.

the integrating factor is e^-3t

Reply 8

OP

Original post by TeeEm

the integrating factor is e^-3t

Thank you my working out shows the answer but I don't understand how my integrating factor is e^-3t

^ds/_dt- 3s = te^3t

This is what I did the first time around can you explain where I'm going wrong with my IF. Please and Thank you.

I(t)= -3 U(t)= e^I(t)= e^-3

Reply 9

if dy/dx + yP(x) = Q(x)

the the integrating factor is always

e to the power of the integral of P(x) ,

Reply 10

OP

Thank you I realise that I've just forgotten to integrate this whole time sorry.

Reply 11

OP

In case any one wants the correct working out to find the general solution in terms of t.
^ds/_dt - 3s = te^3t
P(t)= -3
I(t)= -3t
U(t)= e^-3t
Q(t)=te^3t

^d/_dt(e^-3ts) = te^3t_*e^-3t
^d/_dt(e^-3ts) = t
e^-3ts = ¹/₂t²+ K
s=(¹/₂t²+ K)e^3t

Reply 12