# Problem with initial value first order differential equation.

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.

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

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

• 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. – Antoine Jan 16 at 20:17

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?

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 $$y’ + y = 1+t\implies y=t+ c_1 e^{-t} \qquad \text{with} \qquad c_1=-\frac{3 }{2}e^{3/2}$$

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