Question

Solve the following initial value problems:
y'+2y = 2+e^x; y(0)=4
thnx in advance :)

Answer 1

Let's summon the Mimi....
@Mimi_x3

Answer 2

mimi i choose you :P

Answer 3

okay, in lieu of Mimi, you'll have to work with me :P

Answer 4

ok :D sounds good

Answer 5

Any idea what method we're supposed to be using?

Answer 6

i think we have to find C

Answer 7

then subtitute into the simplified equation, but i still cant quite get it xD

Answer 8

First we need to solve this differential equation :D

Answer 9

ok lets start :)

Answer 10

This needs what's called an integrating factor.

Answer 11

Pikachu's here....
@Mimi_x3 ?

Answer 12

xD

Answer 13

Okay, let's write that equation down...
\[\Large y' +2y = 2+e^x\]

Answer 14

Whenever you have a differential equation of the form...
\[\Large y' + p(\color{red}x)\color{blue}y=q(\color{red}x) \]
You need to find what's called an integrating factor.

Answer 15

What you do is take this part...
\[\Large y' + \boxed{p(\color{red}x)}\color{blue}y=q(\color{red}x)\]
and integrate it.

Answer 16

the 2?

Answer 17

Very good!
\[\Large y' +\boxed2y = 2+e^x\]
so I want you to find
\[\Large \int 2 \ dx\]
Shouldn't be too hard, really :D

Answer 18

so it would be 2y^2/2?

Answer 19

huh?
No, first the integral, can you solve it? :D
\[\Large \int 2 \ dx\]

Answer 20

2x

Answer 21

Okay. So the integrating factor is
\[\Large e^{2x}\]

Answer 22

We multiply this to the entire expression...
\[\Large e^{2x}\frac{dy}{dx}+2e^{2x}y=2e^{2x}+e^{3x}\]

Catch me so far?

Answer 23

yes, but why did we do that? sorry for the dumb question

Answer 24

Very good question.
(not a dumb one ^.^)

Answer 25

You know implicit differentiation?

Answer 26

hmmm took it last sem but i kinda forgot, need a reminder :/

Answer 27

Okay, you notice, if we differentiate this implicitly
\[\Large y e^{2x}\]
Using the product and chain rules, we get
\[\Large e^{2x}y'+2e^{2x}y\]

Right?

Answer 28

hmm

Answer 29

oh yea ok got it

Answer 30

Well, that's what our left-side is now.
\[\Large \color{red}{e^{2x}\frac{dy}{dx}+2e^{2x}y}=2e^{2x}+e^{3x}\]

Answer 31

y' = dy/dx yep

Answer 32

Okay, so the entire left side may NOW be written
\[\Large \frac{d}{dx}(ye^{2x})=2e^{2x}+e^{3x}\]

Answer 33

ok

Answer 34

you may continue :3 if you want

Answer 35

Sorry. It's now basically a separable differential equation.
\[\Large d(ye^{2x}) =( 2e^{2x}+e^{3x})dx\]

Answer 36

yea we moved the dx to the RHS :) nice

Answer 37

we intigrate now?

Answer 38

Yup. Left side just integrates simply.
\[\Large ye^{2x} =\int( 2e^{2x}+e^{3x})dx\]
right side... you do that.

Answer 39

we integrate 2e^2x then e^3x... (2e^2x+1/2) + (e^3x+1/2) + C ? gah i think its wrong :/

Answer 40

Just relax... deep breaths and try again :)

Answer 41

2e^2x+1 is that right? O.o

Answer 42

ahaha... this is an exponential, not a polynomial :D

Answer 43

hmmmm, where did i go wrong on my prev ans xD

Answer 44

Remember that
\[\Large \int e^xdx = e^x+C\]

Answer 45

ooh yea

Answer 46

sry i forgot that one

Answer 47

So redo it :)

Answer 48

so its the same we just add +c to the end?

Answer 49

Not simply. It requires a little substitution.

Answer 50

am out of ideas...

Answer 51

You really should practice Integrals... THEY ARE ESSENTIAL to differential equations :D

Answer 52

I'll integrate for you this once :D
\[\Large ye^{2x} =e^{2x}+\frac13e^{3x}+C\]

Answer 53

So we need to find C.

Answer 54

we use x = 0 and y = 4 right?

Answer 55

Yup.

Answer 56

do we need to simplify further or just substitute?

Answer 57

Just substitute.
\[\Large 4(e^{2(0)}) = e^{2(0)}+\frac13e^{3(0)}+C\]

And solve for C.

Answer 58

4= 1.3+c

Answer 59

No, don't use decimals, use fractions.

Answer 60

2.7 = c?

Answer 61

lol sry

Answer 62

and it's
4 = 1 + (1/3) + C

Answer 63

(1/3) is not 0.3

Answer 64

o_o ok

Answer 65

i think i should change my major

Answer 66

So, just solve this
\[\Large 4 = 1+\frac13 + C\]

I have to go for now.

Answer 67

I'll be back in 20 minutes, maybe :D

Answer 68

ok take your time

Answer 69

aha ok thank you

Answer 70

c = 8/3

Answer 71

its ok, thanks for the suggestion

Answer 72

haha why are you worrying too much? isn't this site for multiple opinions?

Answer 73

There's virtually no such thing as opinions in Maths.
Either I'm correct or I'm wrong.

Answer 74

there are ways to get to the answer right?

Answer 75

Surely.

Answer 76

oh and btw sry i meant c= 16/3 just a little miscalculation

Answer 77

then we subtitute into the equation's C with 16/3 right?