Question

Find solutions of the given IVP.
(1+e^t)dy/dt+e^(t)y=0 y(0)=1

Answer 1

hint : \(\large \left[(1+e^t)y\right]'\)

Answer 2

what

Answer 3

remembe the product rule ?

Answer 4

yea

Answer 5

\(\large (f*g)' = ?\)

Answer 6

f'g+fg'

Answer 7

expand the derivative of product in hint above, what do you get ?

Answer 8

e^t(y')+(1+e^t)y

Answer 9

try again

Answer 10

i dont see a mistake

Answer 11

sure ?

Answer 12

it seems you're doing f'g' + fg

Answer 13

instead of fg' + f'g

Answer 14

oh e^t(y)+(1+e^t)(y')

Answer 15

yes! isn't that same as the left hand side of your differential equation ?

Answer 16

yes

Answer 17

\[\large (1+e^t)dy/dt+e^ty=0 \]

using product rule, becomes :

\[\large \left[(1+e^t)y\right]' = 0\]

Answer 18

which function has "0" as derivative ?

Answer 19

y?

Answer 20

derivative of a constant function is 0

Answer 21

because a constant function never changes

Answer 22

\[\large \left[(1+e^t)y\right]' = 0\]

or you could integrate both sides so that the derivative disappears on left hand side :

\[\large \int \left[(1+e^t)y\right]' dt = \int 0 ~dt\]

\[\large (1+e^t)y = C\]

Answer 23

plugin the initial condition and solve C

Answer 24

plug in 1 for y

Answer 25

yes 1 for y and 0 for t

Answer 26

c=2

Answer 27

yes plug that back into the final equation :

\[\large (1+e^t)y = 2\]

Answer 28

or \(\large y = \dfrac{2}{1+e^t}\) is the solution function for the differential equation

Answer 29

thanks

Answer 30

how would i know which part of the equation to equal it to 0?

Answer 31

what variables do you see in the given equation ?

Answer 32

t,y', and y

Answer 33

thats true, lets keep y' aside.

`t` and `y` are the variables and you're trying to solve for the function `y` right ?

Answer 34

and `y` is a function of `t`

Answer 35

y(0) = 1

means

when t = 0, the value of y is 1

Answer 36

\(\large \large y = \dfrac{2}{1+e^t}\)

assumption is that y is a function of t :

\(\large y(t) = \dfrac{2}{1+e^t}\)

Answer 37

I think of it as initial time idk if that'll help though.

Answer 38

so get t and y' together to equal 0

Answer 39

y(0)=1

you're asking how to interpret this, right ?

Answer 40

im asking why you chose (1+e^t)y'=0 and not e^t y

Answer 41

when did we choose that ?

Answer 42

we never set (1+e^t)y'=0

Answer 43

at the beginning using the product rule

Answer 44

Your starting equation :
\(\large (1+e^t)dy/dt+e^{t}y=0 \)

\(\large (1+e^t)y'+e^{t}y=0 \)

Answer 45

now wrap that using the product rule in reverse

Answer 46

entire left hand side can be compressed as :

\(\large \left[(1+e^t)y\right]'\)

Answer 47

yes ?

Answer 48

why? where did e^t y go

Answer 49

the derivative is NOT just to the y,
it was to the entire expression inside square parenthesis

Answer 50

it did not go anywhere,
expand it using product rule

Answer 51

i thought product rule was just for (1+e^t) and y'

Answer 52

\(\large \large \left[(1+e^t)y\right]'\)

is exactly same as :

\(\large (1+e^t)y'+e^{t}y\)

Answer 53

oh i didn't know that

Answer 54

work it using product rule and convinces yourself they are same

Answer 55

\(\large \large \left[(1+e^t)y\right]' = ?\)

Answer 56

it equals the same thing, so how can I work backwards to obtain that expression?

Answer 57

you need to *see* it and think in reverse

Answer 58

oh thats kinda hard

Answer 59

\(\large \large (1+e^t)y'+e^{t}y\)

there is no rule for writing it as derivative of product without thinking in reverse

Answer 60

however you can convince urself more that it can be written asderivative of product like this :

\(\large (1+e^t)y'+e^{t}y\)

derivative of the first factor in first term \((1+e^t) \) comes as first factor in second term. So thats a huge hint that you can wrap it as : \(\large (fg)'\)

Answer 61

so in all of these kinds of problems I need to do the same thing, like think about what expression would equal that when using the product rule

Answer 62

also notice that one of the functions is always \(\large y\). its not hard at all after you practice 2-3 problems

Answer 63

if you want a practice example, try to write below as derivative of product :

\(\large x^2y' + 2xy\)

Answer 64

[(x^2)y]'

Answer 65

perfect ! want another example ?

Answer 66

so for dy/dt+2ty=t^3

Answer 67

that won't work this simple

Answer 68

as left side cannot be written as derivative of product

Answer 69

the good news is we can force it to be written as derivative of product

Answer 70

multiply something bothsides such that the lefthand side becomes derivative of product

Answer 71

there is a procedure for finidng that "something"

Answer 72

heard of "integrating factor" before ?

Answer 73

yea i think

Answer 74

what do you know about it ?

Answer 75

is it using this \[\mu=e ^{\int\limits_{}^{}p(t)dt}\]

Answer 76

exactly! find it

Answer 77

and multiply that both sides of the equation

Answer 78

so do I solve for y or c

Answer 79

okay i got y=1/2t^2 - 1/2 +c

Answer 80

whats your integrating factor ?

Answer 81

dy/dt+2ty=t^3

Answer 82

integrating factor = \(\large e^{\int 2t~dt} = e^{t^2}\)

right ?

Answer 83

yea

Answer 84

multiply that through out the equation

Answer 85

\(\large dy/dt+2ty=t^3\)

multiiply the integrating factor \(\large \color{gray}{e^{t^2}}\) through out, you get :

\(\large \color{gray}{e^{t^2}}dy/dt+\color{gray}{e^{t^2}}2ty=\color{gray}{e^{t^2}}t^3\)

Answer 86

now, you can blindly write left hand side as \(\large \left[y \color{gray}{e^{t^2} }\right]' \)'

Answer 87

yes

Answer 88

\(\large \left[y \color{gray}{e^{t^2} }\right]' =\color{gray}{e^{t^2}}t^3 \)

Answer 89

integrate both sides

Answer 90

\[\large \int \left[y \color{gray}{e^{t^2} }\right]' ~dt = \int ~\color{gray}{e^{t^2}}t^3~dt \]

Answer 91

left hand side, integral and derivative eat eachother out giving you :

\[\large y \color{gray}{e^{t^2} } = \int ~\color{gray}{e^{t^2}}t^3~dt \]

Answer 92

you will have to evaluate the right hand side integral

Answer 93

i know the integration but how can i find it by hand, product rule?