anonymous
  • anonymous
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
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
is this a separable function?
Jhannybean
  • Jhannybean
I was just about to ask you that :P
Jhannybean
  • Jhannybean
a separable equation is when you can separate the y's from the x's. Can you do that in this equation?

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More answers

anonymous
  • 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
  • anonymous
right now, i solved for C as -1/2 so i have y^2 +y = (x^2-1)/2
Jhannybean
  • 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
  • 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
  • anonymous
haha there was a big jump from where i was at to the answer
dan815
  • dan815
YO BOB can u do this integration thing
Jhannybean
  • Jhannybean
\[\large \int\limits\limits (1+2y)dy = \int\limits\limits xdx\]\[\large y +y^2 =\frac{x^2}{2}+ c\]
dan815
  • dan815
rOB did u learn about ordinary differential equations and stuff yet??
anonymous
  • anonymous
not sure how to do the algebra to solve for y from \[y^{2}+y = \frac{ x^{2} -2}{ 2 } \]
dan815
  • dan815
i have very important life advice to give you.. so u dont make the same mistake as me
anonymous
  • anonymous
yes i know how to do most ODE's, this one was tricky for me
Jhannybean
  • Jhannybean
hint: it's a quadratic.
dan815
  • 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
  • Jhannybean
y(x)=0 y(-1) = 0
Jhannybean
  • Jhannybean
Never done laplace transformations T_T
Jhannybean
  • Jhannybean
Dan can you help me solve this problem....
dan815
  • dan815
wut problem
anonymous
  • anonymous
do i have to use the quadratic equation for this?? ughhh
Jhannybean
  • 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
  • anonymous
i believe the initial values were only there to solve for C
Jhannybean
  • Jhannybean
This is where i have to ask @FutureMathProfessor for guidance... -_-
anonymous
  • anonymous
i don't think we're just allowed to plug in -1 for x, and not 0 for y
Jhannybean
  • Jhannybean
ohh. haha you're probably right.
dan815
  • dan815
umm plug in y value >_>
Jhannybean
  • Jhannybean
xD
anonymous
  • 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
  • Jhannybean
\[\large y+y^2 = \frac{(-1)^2}{2}+ c\]\[\large 0+0^2 = \frac12 + c\]\[c=-\frac12\]
Jhannybean
  • Jhannybean
Soooo.... \[\large y +y^2 =\frac{x^2}{2}+ c\]\[\large y +y^2 =\frac{x^2}{2} - \frac12\]
anonymous
  • 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
  • dan815
yes good are ur done
Jhannybean
  • Jhannybean
yay!!...
anonymous
  • anonymous
now i'm just trying to solve that for y
anonymous
  • anonymous
which i think i have to use the quadratic equation
Jhannybean
  • Jhannybean
Thanks @FutureMathProfessor .
Jhannybean
  • Jhannybean
I don't think you have to solve any more....
dan815
  • dan815
u crazy man!
dan815
  • 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
  • dan815
infinite solutions as f(x) can be any arb function of x
dan815
  • 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
  • anonymous
yeaaaaahhhhhhh......... haha
anonymous
  • 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
  • anonymous
cool, thanks for writing it all out and confirming!

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