Differential equations (initial-value problem)

[DivGradCurl]

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]

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.

 

[DivGradCurl]

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.

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

Thanks

 

[DivGradCurl]

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!

 

[kim.kelly21]

this helped me soooo much! thanks you