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

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

is this a separable function?

- Jhannybean

I was just about to ask you that :P

- Jhannybean

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

right now, i solved for C as -1/2
so i have y^2 +y = (x^2-1)/2

- Jhannybean

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?:)

- anonymous

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

- anonymous

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

- dan815

YO BOB can u do this integration thing

- Jhannybean

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

- dan815

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

- anonymous

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

- dan815

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

- anonymous

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

- Jhannybean

hint: it's a quadratic.

- dan815

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!!!

- Jhannybean

y(x)=0
y(-1) = 0

- Jhannybean

Never done laplace transformations T_T

- Jhannybean

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

- dan815

wut problem

- anonymous

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

- Jhannybean

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

- anonymous

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

- Jhannybean

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

- anonymous

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

- Jhannybean

ohh. haha you're probably right.

- dan815

umm plug in y value >_>

- Jhannybean

xD

- anonymous

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.

- Jhannybean

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

- Jhannybean

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

- anonymous

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

- dan815

yes good are ur done

- Jhannybean

yay!!...

- anonymous

now i'm just trying to solve that for y

- anonymous

which i think i have to use the quadratic equation

- Jhannybean

Thanks @FutureMathProfessor .

- Jhannybean

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

- dan815

u crazy man!

- dan815

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

- dan815

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

- dan815

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...

- anonymous

yeaaaaahhhhhhh......... haha

- anonymous

\[\frac{ dy }{ dt } = \frac{ t }{ (1+2y) }\] 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)

- anonymous

cool, thanks for writing it all out and confirming!

Looking for something else?

Not the answer you are looking for? Search for more explanations.