Question

anyone here familiar with differential equations?

Answer 1

Yes, many of us here are familiar with differential equations.

Answer 2

i wanted to verify a few answers to my homework

Answer 3

there's 7 questions...and i got 5 of the answers...2 i'm absolutely stuck on

Answer 4

find the values of the parameter r for which y = e^rx is a solution of the equations y''+y'-2y = 0

Answer 5

i got r = -2 and r = 1

Answer 6

correct

Answer 7

Just solve the associated auxiliary equation \(r^2+r-2=0 \implies (r+2)(r-1)=0 \implies r=-2\) or \(r=1\).

Answer 8

okay...next one....find the solution of the diferential equation y' = (3x^2)/(1+x^3) such that y(0) = 0

Answer 9

thats right.

Answer 10

i get y = ln(1+x^3)

Answer 11

auxiliary equation is m^2+m-2 = 0
solution -2 and 1 gives your answer

Answer 12

so you integrated it and used the condition, correct? if the procedure is correct, you don't need to worry about the result.

Answer 13

yes that's what i did

Answer 14

Integrate both sides: \(y=\ln\|1+x^3|+c\). Now use the initial value to find \(c\). You will have \(0=\ln(1)+c \implies c=0\). Hence the solution of the DE is \(y=\ln|1+x^3|\).

Answer 15

next one is y'=3x^2e^(-x^3), such that y(0)=1

Answer 16

i get y = e^(-x^3)-2

Answer 17

Not really. If you integrate both sides you get \(y=-e^{-x^3}+c\). Use the initial value to get the value of \(c\).

Answer 18

right that's what i did...c comes out to be 2

Answer 19

so i moved the - in front of e to get e^(-x^3)-2

Answer 20

or should i just leave it alone to be -e^(-x^3)+2?

Answer 21

Are you sure \(c\) is \(2\)?

Answer 22

lol..i thought i was...now i'm not so sure with your reaction lol

Answer 23

If you plug the initial value, you get \(1=1+c \implies c=0\).

Answer 24

:)

Answer 25

isn't x 0?

Answer 26

y(0)=1

Answer 27

Whoops! Sorry :(

Answer 28

\(1=-1+c \implies c=2\). So your answer should be \(y=-e^{-x^3}+2\). Right?! :D

Answer 29

does it matter if i move the -?

Answer 30

like e^(-x^3)-2?

Answer 31

or does it have to remain +2?

Answer 32

Of course it does. That's actually the negative of the value. You can write it as \(y=2-e^{-x^3}\).

Answer 33

okay..yea i started wondering if that changed the entire equation...so yea i get that

Answer 34

this one has me stumped bad....it's y'=3x^2 sqrt(1-y^2) general solution

Answer 35

You can solve it by separating variables. Write the equation as \(\frac{dy}{dx}=3x^2\sqrt{1-y^2}\). By multiplying both sides by \(dx\) and dividing by \(\sqrt{1-y^2}\), we have:
\(\frac{dy}{\sqrt{1-y^2}}=3x^2dx \implies \sin^{-1}y=x^3+c \implies y=\sin(x^3+c)\). You can find \(c\) if you've got a given point of y(x).

Answer 36

wow...you um....made that look really easy lol..

Answer 37

Just to make it clearer, I just put the y's in one side and the x's on the other side and then integrated both sides.

Answer 38

I am glad you found it easy :)

Answer 39

yea i know how to do separating...i just didn't see how to separate til you did it

Answer 40

I gotta go now. I might come back later.

Answer 41

u mind helping me with one other one?

Answer 42

oh okay..thx for your help

Answer 43

Maybe later :)