0
$\begingroup$

I am currently working on a 1ODE problem that states to find the solution to $ y’ + ty = 1+t$ with $ y(3/2) = 0 $ When I try to solve this I get intergral of $e^{{t^2}/2} dt $ somewhere along the line which I tried on several calculators to no avail, unless im missing something. This intergral makes the whole problem hard to complete without long expressions keeping the intergral intact.

Even if I kept it as is, I dont see how to solve the problem by finding an answer for c, or atleast a decently simple answer for c.

Sorry if this isnt very clear, I will edit if needed. Thank you very much.

$\endgroup$
  • $\begingroup$ You can leave $\int e^{t^2/2}dt$ as is, since there is no representation in terms of elementary functions. $\endgroup$ – Shubham Johri Jan 16 at 19:49
0
$\begingroup$

You get$$d(ye^{t^2/2})=(1+t)e^{t^2/2}dt$$Integrate from $3/2$ to $t$.$$\int_{t=3/2}^{t=t}d(ye^{t^2/2})=\int_{3/2}^t(1+t)e^{t^2/2}dt\\y(t)e^{t^2/2}-0=\int_{3/2}^te^{t^2/2}dt+e^{t^2/2}-e^{9/8}$$

$\endgroup$
  • $\begingroup$ Thanks, got it. Was stuck using another method our prof told us to use that simply made things more confusing. I reviewed the textbook along with your answer and it made sense. $\endgroup$ – Antoine Jan 16 at 20:17
0
$\begingroup$

Computing $$\mu=e^{\int t dt}=e^{\frac{t^2}{2}}$$ we get $$e^{t^2/2}y'(t)+e^{t^2/2}y(t)=-e^{t^2/2}(-t-1)$$ and this is $$\frac{d}{dt}\left(e^{t^2/2}y(t)\right)=e^{t^2/2}(t+1)$$ Can you finish?

$\endgroup$
0
$\begingroup$

As you very correctly did $$y'+t y=0 \implies y=C\, e^{-\frac{t^2}{2}}$$ Using now the variation of parameters, you end with $$C'\, e^{-\frac{t^2}{2}}=1+t \implies C'=(1+t)\, e^{\frac{t^2}{2}}$$ and the problem leads to special function $$C=c_1+e^{\frac{t^2}{2}}+\sqrt{\frac{\pi }{2}} \text{erfi}\left(\frac{t}{\sqrt{2}}\right)$$ where appears the imaginary error function. So, by the end, $$y=1+\sqrt{\frac{\pi }{2}} e^{-\frac{t^2}{2}} \text{erfi}\left(\frac{t}{\sqrt{2}}\right)+c_1 e^{-\frac{t^2}{2}}$$ Using the condition, we end with $$c_1=-\sqrt{\frac{\pi }{2}} \text{erfi}\left(\frac{3}{2 \sqrt{2}}\right)-e^{9/8}$$ which you could evaluate using the series expansion $$\operatorname{erfi}(z)= \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty\frac{z^{2n+1}}{n! (2n+1)} =\frac{2}{\sqrt{\pi}} \left(z+\frac{z^3}{3}+\frac{z^5}{10}+\frac{z^7}{42}+\frac{z^9}{216}+\cdots\right)$$

Using seven terms in the summation, you should get $\text{erfi}\left(\frac{3}{2 \sqrt{2}}\right)\approx 1.84849$ which is correct for six significant figures and the $c_1\approx -5.39695$.

If you are not supposed to know about this special function,we can assume a typo in the problem and suppose that is was in fact $$y’ + y = 1+t\implies y=t+ c_1 e^{-t} \qquad \text{with} \qquad c_1=-\frac{3 }{2}e^{3/2}$$

$\endgroup$
  • $\begingroup$ Actually I believe that I was just supposed to keep it in terms of the intergral without solving it. Thanks for the breakdown though. $\endgroup$ – Antoine Jan 17 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.