Where do these other initial conditions come from?I have the problem 2dydt=y(y−2),y(0)=3, which I get how to solve up until the intial conditions. The solution is y(t)=21−Cet. The solution then plugs in three initial conditions and solves for C. They do y(0)=1,y(0)=−1, and y(0)=3 and then solve each of these. Where did the other two come from?Also, the problem asks to determine the interval of existence. This one is straightforward, but there are others where the interval is something like (−∞,2)∪(2,∞) and they plug in t=0 and then take only the interval that solution lies in. Why is that?Exact wording: Identify the equilibrium (constant) solutions for this nonlinear ODE. a) identify the equilibrium solutions for the nonlinear ODE. b) Solve the IVP for the initial condition, y(0)=3. Determine the existence interval and the limit of y(t) as t approaches the endpoints of the interval.

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zakulisan337 · Accepted Answer

Step 1The only initial condition given by the problem is y(0)=3, so there is no reason why the other ones should be relevant. Regarding the equilibrium solutions, notice that this equation is of the formy′=f(y).The equilibrium solutions are given by the constant functions y such that f(y)=0. I will leave it up to you to figure it out from here.As for the initial value problem, the equation is equivalent to2y(y−2)y′=1,and notice that 2y(y−2)=y−(y−2)y(y−2)=1y−2−1y,hence y′y−2−y′y=1.Step 2By integrating, one has thatln⁡[y(t)−2y(0)−2]−ln⁡[y(t)y(0)]=t=ln⁡[y(t)−2]−ln⁡[y(t)]+ln⁡(3).As such, y(t)−2y(t)=1−2y(t)=et3.This is equivalent to y(t)=21−et3=63−et.Notice that y has a singularity at ln(3). For t&amp;gt;ln⁡(3), we have that et&amp;gt;3, hence −et&amp;lt;−3, so 3−et&amp;lt;0, which is equivalent to y(t)&amp;lt;0. This contradicts the condition that y(t)&amp;gt;2, and this is why only the interval (−∞,ln⁡(3)) is taken as the domain of the solution. The initial condition y(0)=3 does not determine a unique solution of the differential equation on (ln⁡(3),∞).

Summer Berg · Accepted Answer

Explanation:Given the Cauchy problem{2y′=y(y−2)y(0)=ayou can find the equilibrium solutions y≡0 for a= and y≡2 for a=2. Since f(x,y)=12y(y−2)∈C1(R2), we know that f is continuous and locally lipshitz in y uniformly respect to x, which means that ∃d,δ,L&amp;gt;0 such that x∈[−d,d],y1,y2∈B―δ(a)⇒Vertf(x,y1)−f(x,y2)Vert≤Verty1−y2Vert. This implies that ∀a∈R,∃Ia⊂R, with 0∈Ia, such that it&#039;s defined a unique solution for the Cauchy problem ya:Ia→R. For the theorem of existence and uniqueness, the solutions ya, for a≠0,2, must be contained in the region (−∞,0) or (0, 2) or (2,+∞) (the solutions defined for a∈R∖{0,2} can&#039;t intersect the functions y≡0,2).Assuming a≠0,2 you can solve the equation∫ay2dζζ(ζ+2)=∫0xdξ⇒∫ay2ζ(ζ+2)dζ=x solog⁡(dyy−2)−log⁡(daa−2)=x and from this relation we can find y:log⁡(yy−2)=x+log⁡(aa−2)∖yy−2=ex+log⁡(aa−2)∖y=ex+log⁡(aa−2)(y−2)∖y(x)=d−2ex+log⁡(aa−2)1−ex+log⁡(aa−2)=d−2aa−2ex1−aa−2exThe interval Ia is determined by the parameter a.

Where do these other initial conditions come from? I

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