2
$\begingroup$

From Logan, D. (2017). A first course in differential equations. Springer.enter image description here

enter image description here

What are the steps to simplify that derivative?

EDIT: As pointed out in the comments there is likely a typo in the answer to 11. With that in mind it is actually quite simple to solve this. As I have done below. The initial derivative of $y(t)$ should read....

$$ y'(t) = e^{-t^2}e^{t^2}-2te^{-t^2} \int_0^t e^{s^2}ds $$

$\endgroup$
5
  • 6
    $\begingroup$ Could there be a typo in 11, i.e., an $e^{-t^2}$ accidentally became $e-t^2$? No one can tell except you by actually using the product rule as suggested … $\endgroup$
    – Hagen von Eitzen
    Commented Jan 4, 2022 at 8:05
  • 1
    $\begingroup$ If you wrote the equations in MathJax, the question would look much better :) $\endgroup$
    – Hermis14
    Commented Jan 4, 2022 at 8:20
  • 2
    $\begingroup$ @Hermis14 Then again, in that case we would have suspected that The typo was introduced by Angus and not the original $\endgroup$
    – Hagen von Eitzen
    Commented Jan 4, 2022 at 9:06
  • $\begingroup$ Yes thank you now it is quite simple to resolve the steps taken. I was very confused with the initial statement and assumed I had done something horribly wrong. $\endgroup$
    – Angus Campbell
    Commented Jan 4, 2022 at 17:47
  • 1
    $\begingroup$ Please, use descriptive titles. "Understanding the following steps..." says nothing about the subject of the question. $\endgroup$
    – jjagmath
    Commented Jan 4, 2022 at 19:28

1 Answer 1

Reset to default
0
$\begingroup$

As pointed out in the comments there is likely a typo in the answer to 11. With that in mind it is actually quite simple to solve this. So taking the initial derivative we apply the product rule and the FTC1 to get...

$$ y'(t) = e^{-t^2}e^{t^2}-2te^{-t^2} \int_0^t e^{s^2}ds $$ $$ y'(t) = \frac{e^{t^2}}{e^{t^2}}-2t\underbrace{e^{-t^2} \int_0^t e^{s^2}ds}_{=y} $$ $$ y'(t) = 1- 2ty$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .