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I want to find a first order ode, an initial value problem, that has the solution $$y=(1-y_0)t+y_0$$ where $y_0$ is the initial value.The ode has to be of first order, that is: $$y'=f(y).$$ I need this to test a special solver I am building. The main objective is to find an ode that has the property that the end-value goes into the reverse direction of the initial value. My thought is, that the function above is the most simple one that fulfills that requirement. However, any other idea that produces my desired result is welcome.

I came so far: since $$y'=1-y_0=f((1-y_0)+y_0)$$ $f$ should be something like $$f(z)=\frac{z-y_0}{t}$$ so we get: $$y'=\frac{y-y_0}{t}$$ However, I don't know how to properly get the initial condition into the equation, the $y_0$ part in the ode itself doesn't seem right, since the initial value can't be put into the ode itself but is a special constrained outside the ode. Can anyone help me get this clear?

2 Answers 2

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Solve for $y_0$:

$$ y_0 = \frac{t-y}{t-1} $$

Then differentiate: $$ 0 = \frac{(t-1)(1-y') - (t-y)}{(t-1)^2} = - \frac{y'}{t-1} + \frac{y-1}{(t-1)^2}$$ i.e. $$ y' = \frac{y-1}{t-1}$$

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    Your $f$ is not of the required form. – 2017-02-19
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    Your method delivers the correct ode to my question, thanks. But I don't get what you're doing. Could you please elaborate on that a little? Why are you doing step one and step two? – 2017-02-20
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    You don't want the constant $y_0$ in your equation. A way to get rid of it is to differentiate it. – 2017-02-20
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    cool idea, thanks – 2017-02-21
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This is maybe not what you intended.

Note that you want a differential equation of the form $y'=f(y)$, and at the same time the proposed solution has constant derivative with respect to $t$. This implies that $f$ is necessarily of the form $f(y)=c$ for some constant $c$. Given $y_0$ an IVP satisfying your requirements therefore could be $$y'=1-y_0,\qquad y(0)=y_0\ .$$

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    This is simple, however I first would like to have a real ode (not a trivial case), and second, you have $y_0$ in your equation, which cannot be allowed, since I want the same equation with different initial values, instead of a new equation each time. – 2017-02-20