# DSolve with assumptions

I'm new to mathematica and was wondering if someone could help me. I have a differential equation which is:

$\frac{dy}{dx} = 1/2 y(1 - y)(a - y + a y)$

It has initial condition $y(0) = y0$ and it's assumed $0 < a < 1/2$.

So I've been trying to figure out how to use DSolve. I know the solution $y(x)$ is meant to fall in the range $0 < y(x) < 1$ but DSolve keeps giving me complex solutions. I'm assuming this is because I haven't provided it any assumptions. Here's what I've tried:

DSolve[{y'[x] ==
1/2 (1 - y[x]) y[x] (a - y[x] + a y[x]), y[0] == y0}, y[x], x]


Which results in:

{{y -> Function[{x},
InverseFunction[(
a Log[1 - #1] + (1 - 2 a) Log[#1] + (-1 + a) Log[
a - #1 + a #1])/(a (-1 + 2 a)) &][-(x/2) + (
a Log[1 - y0] + Log[y0] - 2 a Log[y0] - Log[a - y0 + a y0] +
a Log[a - y0 + a y0])/(a (-1 + 2 a))]]}}


There are a few questions I was wondering someone could help me with:

• Is it possible to pass assumptions to DSolve, e.g. to tell it $0 < a < 1/2$ and $0 \leq y \leq 1$? If not how can I handle this in Mathematica
• What is the correct ways of assigning the result of DSolve to a function whereby I can pass values of $a$, $x_0$ and $x$, e.g. $y(x, x_0, x)$?
• I can add this about why M can't solve the InverseFunction problem: It's similar to http://mathematica.stackexchange.com/q/56771, but the exponential form of your equation has terms of the form y^2 and y^a. For a generic a, that's hard to solve, unless you're lucky. Commented Apr 26, 2015 at 19:44

To create a function of the type you ask for, you can do the following. Note the use of Set and DSolveValue.

Clear[a, y0, x];
y[a_, y0_, x_] =
DSolveValue[{y'[x] == 1/2 (1 - y[x]) y[x] (a - y[x] + a y[x]),
y[0] == y0}, y[x], x];

Plot[Evaluate@Table[y[a, 0.1, x], {a, 0.1, 0.4, 0.1}], {x, 0, 50}]


Instead of DSolveValue one can also use y[x] /. DSolve[<ode>, y[x], x].

One can pass assumptions to the functions used by DSolve via $Assumptions, which can be temporarily set with Assuming. That sometimes helps, but not in this case. Assuming[0 < a < 1/2 && 0 < y[x] < 1 && 0 < y0 < 1, DSolveValue[{y'[x] == 1/2 (1 - y[x]) y[x] (a - y[x] + a y[x]), y[0] == y0}, y[x], x] ]  It does not help in this case because the equation that implicitly defines the solution y[x] cannot be solved. Assuming[0 < a < 1/2 && 0 <= y[x] <= 1 && 0 <= y0 <= 1, eq = DSolveValue[{y'[x] == 1/2 (1 - y[x]) y[x] (a - y[x] + a y[x]), y[1] == y0}, y[x], x] /. InverseFunction[f_][f0_] :> f[y] == f0 // Simplify ] (* a + 2 a^2 x + 2 a Log[1 - y] + (2 - 4 a) Log[y] + 2 a Log[a - y + a y] + 4 a Log[y0] + 2 Log[a - y0 + a y0] == 2 a^2 + a x + 2 Log[a - y + a y] + 2 a Log[1 - y0] + 2 Log[y0] + 2 a Log[a - y0 + a y0] *) Assuming[0 < a < 1/2 && 0 <= y <= 1 && 0 <= y0 <= 1, Solve[eq == 0, y] ]  Solve::nsmet: This system cannot be solved with the methods available to Solve. >> Another option for you is to use ParametricNDSolve, which numerically solves differential equations with one or more parameters. soln = ParametricNDSolveValue[ {y'[x] == 1/2 (1 - y[x]) y[x] (a - y[x] + a y[x]), y[0] == y0}, (* your equations*) y, (* the expression you want to be returned*) {x, 0, 100}, (* the independent variable and range to solve for *) {{a, 0, 0.5}, {y0, 0, 1}} (* the parameters with their ranges *) ]  ParametricNDSolveValue returns a ParametricFunction. You can call this function with a set of parameters to get y as a function of x: soln[0.2, 0.6] (* returns InterpolatingFunction[{{0., 100.}}, <>] *)  Note that the parameters are input in the same order as they were input to ParametricNDSolve, in this case [a, y0]. The InterpolatingFunction will allow you to find the value of y at a specific value of x: soln[0.2, 0.6][10] (* === 0.381214 *)  Although your given system does seem to have an analytic solution, this method will be useful for systems that require numerical solution. One way you can let Mathematica know what you are assuming here is Simplify[DSolve[{y'[x]==1/2 (1-y[x]) y[x] (a-y[x]+a y[x]), y[0]==y0}, y[x], x], 0<a<1/2]  One way of assigning the result and passing value of a, y0, etc is yf = y[x]/.DSolve[{y'[x]==1/2 (1-y[x]) y[x] (a-y[x]+a y[x]), y[0]==y0}, y[x], x][[1]]; Simplify[yf/.{a->2, y0->1/2, x->4}]  That does warn you that inverse functions are being used and you should check the results carefully, but the result from DSolve is an inverse function, so I don't think that can be avoided • @2012rcampion DSolve calls functions (such as Integrate and Simplify) which do use $Assumptions. So Assuming can make a difference. See mathematica.stackexchange.com/a/65946 or mathematica.stackexchange.com/a/75243, for instance. Commented Apr 26, 2015 at 17:15
• @MichaelE2 I didn't know that... that's actually quite useful. Commented Apr 26, 2015 at 19:14
• @2012rcampion It can be useful. So can Block[{Simplify = FullSimplify}, DSolve[ode, y, x]]. But DSolve works well enough on a generically wide range of problems that these tricks do not help that often. Commented Apr 26, 2015 at 19:19