# Differential equations (initial-value problem)

Consider the initial-value problem

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

Find the value of $$y_0$$ for which the solution touches, but does not cross, the $$t$$-axis.

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

Any help is highly appreciated.

First, I use the method of integrating factors.

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

It follows that

$$p(t)=\frac{2}{3}$$

and

$$g(t)=1 - \frac{1}{2}t$$. Thus, we find

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

and

$$y(t)=\frac{1}{\mu (t)}\left[ \int \mu (t) g (t) \: dt + \mathrm{C} \right]$$
$$y(t)=e^{-2t/3}\left[ \int e^{2t/3} \left( 1 - \frac{1}{2}t \right) \: dt + \mathrm{C} \right]$$
$$y(t)=e^{-2t/3}\left[ \left( \frac{21}{8} - \frac{3}{4}t \right) e^{2t/3} + \mathrm{C} \right]$$
$$y(t)=\frac{21}{8} - \frac{3}{4}t + \mathrm{C} e^{-2t/3}$$

The initial condition gives

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

So, we get

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

The problem now is finding the appropriate value for $$y_0$$. The answer my textbook gives is $$y_0\approx -1.642876$$, 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 $$y_0$$ that gives the line $$y=0$$. But again, I'm not so sure. Am I on the right track?

Thanks!

saltydog
Homework Helper
Hello Thiago. The equation is not a horizontal line. Not sure I understand what you mean by that. How about this: if it's continuous and touches the x-axis but does not cross it (and is not y=0), then aren't y and y' zero there? So that's two equations in two unknowns, albeit non-linear ones. It's a start anyway.

Yeah... you're right! What was I thinking?!? If it were a line, it'd fit the form y = ax + b. I must have overlooked my y. :rofl:

Anyway, I see what you mean. I'll use y = 0 and y' = 0 this time.

Thanks

Last edited:
Hey, thank you so much. That works fine!

First, we look into $$y=0$$.

$$\frac{21}{8} - \frac{3}{4}t + \left( y_0 - \frac{21}{8} \right) e^{-2t/3} = 0$$

Solving it for $$y_0$$ gives $$y_0 = \frac{3}{8} \left( 7 - 7 e^{2t/3} + 2t e^{2t/3} \right)$$. Next, we look into $$y^{\prime}=0$$.

$$-\frac{3}{4} -\frac{2}{3}\left( y_0 - \frac{21}{8} \right) e^{-2t/3} = 0$$

Using the $$y_0$$ found in the last step, and solving for $$t$$ gives $$t=2$$. Finally, we substitute that value back in $$y_0$$ and get $$y_0 = \frac{3}{8}\left( 7 - 3e^{4/3} \right) \approx -1.642876$$

That's it!

this helped me soooo much!! thanks you