1
$\begingroup$

question

question2

Question 3

I am trying to solve the differential equation, but I do not understand the method. Here is my working:

My working

  • 0
    how do you got the second equation in the first line? – 2017-01-03
  • 0
    Please see the updated question I have included all of the context, and I will look at reverse product rule. – 2017-01-03
  • 0
    @Moo can you tell me what line of my working I have gone wrong on? I have also multiplied every term by the I.F. Is it line 5 after I have evaluated the differential? – 2017-01-03
  • 1
    Your mistake is that the integrating factor should make the entire left side be of the form $\frac{d}{dt}(i(t) e^t)$. The first term is obtained when the product rule tells you to differentiate $i(t)$, the second term is obtained when it tells you to differentiate $e^t$. – 2017-01-03

2 Answers 2

1

From the first diagram-

$\frac{di}{dt} + i = 10t.e^{-t}$

Multiply equation by $e^t$

$\frac{di}{dt}.e^t + i.e^t = 10t$

Now $\frac{di}{dt}.e^t + i.e^t = \frac{d}{dt}(i.e^t)$

So we have $\frac{d}{dt}(i.e^t) = 10t$

Now integrate to solve further.

Step 3-

$\frac{d}{dt}(i.e^t) = i.\frac{d}{dt}(e^t) + e^t\frac{di}{dt}(i)$

= $i.e^t + \frac{di}{dt}.e^t$

  • 0
    Thank you, on your line 3 of working I do not understand why that is the case. – 2017-01-03
  • 0
    Let me explain you we have $d/dt(i.e^t)=i.d/dt(e^t)+e^t.d/dt(i)$ which is equal to $i.e^t+e^t.di/dt$ – 2017-01-03
  • 0
    I edited my answer. Both are same thing so we replaced. – 2017-01-03
1

The step from line 3 to line 4 isn't correct: $\frac{di}{dt} e^t + ie^t = \frac{d}{dt}(ie^t)$, so line 4 should read $\frac{d}{dt}(ie^t) = 10t$ which can then be integrated directly.

In general, the ideas behind an integrating factor are outlined here as well as I could explain (if not better).