Question

Find the solution to the initial value problem

y' = x/(1+2y) y(-1) = 0


so when solving the diff eq, i'm a little confused on how to solve first order nonlinear differential equations.

i believe we have to set x/(1+2y) to f(x,y)... but i'm lost after that

Answer 1

is this a separable function?

Answer 2

I was just about to ask you that :P

Answer 3

a separable equation is when you can separate the y's from the x's. Can you do that in this equation?

Answer 4

yeah, whenever i was moving (1+2y) to the left side, i multiplied dy throughout, thinking that it would be a problem using integrating factors

Answer 5

right now, i solved for C as -1/2

so i have y^2 +y = (x^2-1)/2

Answer 6

first ofall,you can rewrite this equation as \[\large \frac{dy}{dx}=\frac{x}{1+2y}\]\[\large (1+2y)dy = xdx\] Integrate \[\large \int\limits (1+2y)dy = \int\limits xdx\] Are you good from there?:)

Answer 7

check this
http://www.wolframalpha.com/input/?i=y%27+%3D+x%2F%281%2B2y%29+&dataset=&equal=Submit
try the step by step solution button

Answer 8

haha there was a big jump from where i was at to the answer

Answer 9

YO BOB can u do this integration thing

Answer 10

\[\large \int\limits\limits (1+2y)dy = \int\limits\limits xdx\]\[\large y +y^2 =\frac{x^2}{2}+ c\]

Answer 11

rOB did u learn about ordinary differential equations and stuff yet??

Answer 12

not sure how to do the algebra to solve for y from \[y^{2}+y = \frac{ x^{2} -2}{ 2 } \]

Answer 13

i have very important life advice to give you.. so u dont make the same mistake as me

Answer 14

yes i know how to do most ODE's, this one was tricky for me

Answer 15

hint: it's a quadratic.

Answer 16

OKAY just listen!! when they teach you have LAPLACE transforms... practice it like MAD!! only DO laplace transform for everything its the most useful thing ever!!!

Answer 17

y(x)=0

y(-1) = 0

Answer 18

Never done laplace transformations T_T

Answer 19

Dan can you help me solve this problem....

Answer 20

wut problem

Answer 21

do i have to use the quadratic equation for this?? ughhh

Answer 22

\[\large y+y^2 = \frac{(-1)^2}{2}+ c\]\[\large y^2 + y -\frac{1}{2} = c\] Im just going off what was given. :\

Answer 23

i believe the initial values were only there to solve for C

Answer 24

This is where i have to ask @FutureMathProfessor for guidance... -_-

Answer 25

i don't think we're just allowed to plug in -1 for x, and not 0 for y

Answer 26

ohh. haha you're probably right.

Answer 27

umm plug in y value >_>

Answer 28

xD

Answer 29

y(-1) = 0
Translation: When you plug in -1 for X in your equation, you WILL END UP GETTING ZERO.
Any offset to zero is the value of your C.

Answer 30

\[\large y+y^2 = \frac{(-1)^2}{2}+ c\]\[\large 0+0^2 = \frac12 + c\]\[c=-\frac12\]

Answer 31

Soooo.... \[\large y +y^2 =\frac{x^2}{2}+ c\]\[\large y +y^2 =\frac{x^2}{2} - \frac12\]

Answer 32

i had y^2 +y = (1/2)x^2 + C

so when i plugged in y(-1) = 0, i got -1/2 for C

Answer 33

yes good are ur done

Answer 34

yay!!...

Answer 35

now i'm just trying to solve that for y

Answer 36

which i think i have to use the quadratic equation

Answer 37

Thanks @FutureMathProfessor .

Answer 38

I don't think you have to solve any more....

Answer 39

u crazy man!

Answer 40

basically your solution tells you that you can have any function of x --> f(x) + the f(x)^2 = x^2/2 - 1/2

Answer 41

infinite solutions as f(x) can be any arb function of x

Answer 42

dont worry if that doesnt make sense right now.. its gonna make complete sense as you do more differential equations and when u understand what exactly are these solutions u are finding...

Answer 43

yeaaaaahhhhhhh......... haha

Answer 44

\[\frac{ dy }{ dt } = \frac{ t }{ (1+2y) }\] Then
\[(1 + 2y) dy = tdt\]Then
\[\int\limits_{}^{}(1 + 2y) dy =\int\limits_{}^{} tdt\]Then
\[y+y^2+C1 = \frac{ t^2 }{ 2 }+C2\]Now move over constants to one side as both are arbitrary (We'll move them to the T side, as customary)
\[y+y^2+= \frac{ t^2 }{ 2 }+C\]
Now we can look at our initial condition to get some equal equations so we can find our determined value of C.
at time = -1, our left side will equal zero, so we have:\[y+y^2+= \frac{ (-1)^2 }{ 2 }+C = 0\]
Now since we have an equation \[\frac{ (-1)^2 }{ 2 } + C = 0\]
We merely solve for C, and we will get C = -.5

So our initial value problem is:\[y+y^2= \frac{ t^2 }{ 2 }+\frac{ 1 }{ 2 }\]
We can confirm this by taking the derivative of this function with respect to Y on the left and T on the right and we will get the differential equation that we started on.

(I promise you, when I took my first exam in differential equations, I failed it, so don't call me smart for knowing this)

Answer 45

cool, thanks for writing it all out and confirming!