y' + 7/2y = 1 - 1/7t y(0) = y0
Find the value for yo for which the solution touches, but does not cross, the t-axis. I understand that you solve for y0 and substitute it back into the equation.
I get: y0 = 0.30 + 8c/7. What do I do next?
y' + 7/2y = 1 - 1/7t y(0) = y0
Find the value for yo for which the solution touches, but does not cross, the t-axis. I understand that you solve for y0 and substitute it back into the equation.
I get: y0 = 0.30 + 8c/7. What do I do next?
So I walk into class and Professor Bob writes on the board:
$\displaystyle y'+7/2y=1-1/7t,\quad y(0)=y_0$
and says "Solve this. Ok that's it," and then walks out.
Some of the smart ones (not me) walk up to his office and ask for clarification. But he's eccentric and doesn't give any. What do I do?
Consider first:
$\displaystyle y'+\frac{7}{2} y=1-\frac{1}{7} t,\quad y(0)=y_0$
That's doable via integration factor and has a solution of:
$\displaystyle y_1(t)=\frac{102}{343}-\frac{2}{49}t+k e^{-7/2 t}$
However, it looks to me to eventually cross the t-axis no matter what the initial starting value is.
Next, how about:
$\displaystyle y'+\frac{7}{2y}=1-\frac{1}{7} t$
That's makes it non-linear; our professor is way into that stuff but this is just basic DE. Surely he doesn't want us to solve that one. There goes my A.
Third:
$\displaystyle y'+\frac{7}{2} y=1-\frac{1}{7t}$
I think that one would incur a singularity at t=0. Right? Anyone else teaching this class?
And finally:
$\displaystyle y'+\frac{7}{2y}=1-\frac{1}{7t}$
. . . Mathematica can't even solve that one. Could work on it numerically I suppose. Is Bob gonna' be like this the whole semester?
Next class we go over it, and Professor Bob says, "always try to be extremely precise when you're describing mathematics."