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[tex]y^{\prime} + \frac{2}{3} y = 1 - \frac{1}{2}t,\quad y(0) = y_0[/tex]

Find the value of [tex]y_0[/tex] for which the solution touches, but does not cross, the [tex]t[/tex]-axis.

**The only problem that I seem to have is finding this particular solution that "touches, but does not cross, the [tex]t[/tex]-axis".**

Any help is highly appreciated.

Any help is highly appreciated.

First, I use the method of integrating factors.

[tex]y^{\prime} + \frac{2}{3} y = 1 - \frac{1}{2}t \qquad \mbox{(Standard Form)}[/tex]

It follows that

[tex]p(t)=\frac{2}{3}[/tex]

and

[tex]g(t)=1 - \frac{1}{2}t[/tex]. Thus, we find

[tex]\mu (t) = \exp \int p(t) \: dt = \exp \frac{2}{3} \int \: dt = e^{2t/3}[/tex]

and

[tex]y(t)=\frac{1}{\mu (t)}\left[ \int \mu (t) g (t) \: dt + \mathrm{C} \right][/tex]

[tex]y(t)=e^{-2t/3}\left[ \int e^{2t/3} \left( 1 - \frac{1}{2}t \right) \: dt + \mathrm{C} \right][/tex]

[tex]y(t)=e^{-2t/3}\left[ \left( \frac{21}{8} - \frac{3}{4}t \right) e^{2t/3} + \mathrm{C} \right][/tex]

[tex]y(t)=\frac{21}{8} - \frac{3}{4}t + \mathrm{C} e^{-2t/3}[/tex]

The initial condition gives

[tex]y_0 = \frac{21}{8} - \frac{3}{4}(0) + \mathrm{C} e^{-2(0)/3} \Longrightarrow \mathrm{C} = y_0 - \frac{21}{8}[/tex]

So, we get

[tex]y(t)=\frac{21}{8} - \frac{3}{4}t + \left( y_0 - \frac{21}{8} \right) e^{-2t/3}[/tex]

The problem now is finding the appropriate value for [tex]y_0[/tex]. The answer my textbook gives is [tex]y_0\approx -1.642876[/tex], but so far I haven't been able to figure out the way to work backwards from there. Anyhow, I tried to find it by plotting some of the solutions. It turns out that all of them are horizontal lines. So, maybe I just need to find [tex]y_0[/tex] that gives the line [tex]y=0[/tex]. But again, I'm not so sure. Am I on the right track?

**Thanks!**