Question

how come x^2y'+2xy can be written as (x^2*y)'?? is there something similar we can do with xln(x)y'+y?

Answer 1

find the derivative of x^2y using the product rule for differentiation

Answer 2

i don't understand

Answer 3

the product rule:
\[\frac{ d }{ dx }=f(x)g'(x)+g(x)f'(x)\]

Answer 4

sorry should be
\[\frac{ d }{ dx }(f(x)g(x))=f(x)g'(x)+g(x)f'(x)\]

Answer 5

so the derivative of x^2y = 2xy+y'x^2 OHHHHHHH

Answer 6

yeah i said the same thing when it clicked for me LOL

Answer 7

and no you can't do it for the second one, you'll have to find the integrating factor

Answer 8

nice! is there a similar relation in xln(x)y'+y. (i am assuming there is because all the questions in that section seem to have these patterns)

Answer 9

OK!

Answer 10

well you'll have to find the integrating factor which transforms a linear equation into one that we can solve in that form... can i show you an easier problem than that one to start with, its pretty nasty

Answer 11

wait, you might want the complete question. the question is xln(x)y'+y=xe^x

Answer 12

do you see anything i should be doing in this question

Answer 13

yeah its still nasty lol how about i show you an easy one then get into a nasty one

Answer 14

alright

Answer 15

you need to divide by x(ln(x))

Answer 16

i can show you this one but its gonna be ugly

Answer 17

When we combine the Chain and Product Rules, we get:\[\bf (x^2y)'=(x^2)'y+x^2(y)'=2xy+x^2\frac{ dy }{ dx }\]

Answer 18

@natto

Answer 19

yup i get y'+y/xlnx = xe^x/xlnx

Answer 20

i treid isolating y from x, so i can assign dy and dx to them respectively, but clearly i wasn't able to

Answer 21

ok its in linear form like this

Answer 22

its gonna take a bit for me to type

Answer 23

alright! take ur time!

Answer 24

\[y'+\frac{ 1 }{ xln(x) }y=\frac{ e^{x} }{ \ln(x) }\]

Answer 25

ok what we need is a magic function to multiply both sides of the equation so we can use that fancy product rule. Turns out there is a trick to find this function.

Answer 26

oh nm that integral isn't that bad ha! it just looks bad

Answer 27

ok if the equation is of the form:
\[y'+P(x)y=f(x)\]
then the magic function is found by
\[e ^{\int\limits_{}^{}P(x)dx}\]

Answer 28

can you find
\[\int\limits_{}^{}\frac{ 1 }{ xln(x) }dx\]
(hint: its a u substitution)

Answer 29

yes

Answer 30

ok get that real quick

Answer 31

ln(ln(x))+c

Answer 32

good ok now what is
\[e ^{\ln(\ln(x))}\]

Answer 33

remember inverses (don't worry about abs bars for now, this isn't nitpick class)

Answer 34

e^1

Answer 35

its just ln(x)

Answer 36

remember e^ln(whatever) = whatever

Answer 37

ok!!

Answer 38

Now that we found our magical function, we're gonna multiply both sides of the equation by it, and you will see it become much prettier (i've done this problem before, we must have the same book)

Answer 39

\[(\ln(x))y'+\frac{ 1 }{ x }y=e^{x}\]

Answer 40

the left side is now the product rule... poof majic

Answer 41

see? the left hand side is the derivative of (ln(x))y

Answer 42

omg how do you see that far ahead ?

Answer 43

i've taught differential equations 6 semesters :) you see the pattern one is the derivative of the other

Answer 44

that and its magic... it always works.. these are always solveable

Answer 45

if you multply a derivative in standard linear form by that magic e^integral of P it always gives you a product rule... always

Answer 46

but we're not done we need to solve for y...

Answer 47

Oh! is there a name for that? or more examples that i can up on? i will look it up!

Answer 48

now we see the product rule on the left we have
\[\frac{ d }{ dx }[\ln(x)y]=e^{x}\]

Answer 49

yes?

Answer 50

you said "magic e^integral of P it always gives you a product rule... always" is there a name or keyword for that rule so i can search it up and read on it later?

Answer 51

you need to antidifferentiate both sides using the Fundamental Theorem of Calculus... yes i give you information when we finish the problem :)

Answer 52

ok, working ahead , i get y=e^x/(ln(x))

Answer 53

almost yu forgot the constant

Answer 54

right! +c!! e^x+c/ln(x)

Answer 55

\[y=\frac{ e^{x} }{ \ln(x) }+\frac{C}{\ln(x)}\]

Answer 56

good job... These are called Linear First order Ordinary Differential Equations. They for the most part are solved using the same method. The "majic equation" is called an integrating factor. This is a good video by a professor at MIT here http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/

Answer 57

thank you so much!! you've been super helpful!

Answer 58

np. Differential equations are fun especially when student has a good background in calculus like yourself. :)

Answer 59

ill watch the video and go to sleep :D thanks alot ! good night!