$\begingroup$

Q. Solve each of the following initial-value problems:

$$ 2\frac{d^2y}{dt^2}+\frac{dy}{dt}-10y=0 \text{ where } y(1)=5 \text{ and } y'(1)=2 $$

Now my professor has been giving problems with only y(0) but not with y(1), so I saw this problem in the textbook and I am interested in finding out how to solve them.

So this is my steps so far:

Characteristic equation: $$2r^2+r-10=0$$

I found the roots to be: $$r_1=\frac{-5}{2} \text{ and } r_2=2$$

So I have:

$$ y(t)=c_1 e^{\frac{-5}{2} t} + c_2 e^{2t} $$

but the book says to solve for $c_1$ and $c_2$, my equation needs to be in this form:

$$ y(t)=c_1e^{r_1(t-t_0)}+c_2e^{r_2(t-t_0)} $$

so does that mean I would need to have:

$$ y(t)=c_1 e^{\frac{-5}{2} (t-t_0)} + c_2 e^{2(t-t_0)} $$

and then to solve for the $c_1$ and $c_2$, I would do:

$$c_1+c_2=5$$ $$\frac{-5}{2}c_1+2c_2=2$$

and solve for $c_1$ and $c_2$ by substitution. Does that look right?

$\endgroup$ 1

1 Answer

$\begingroup$

Your form

$$y(t)=c_1 e^{\frac{-5}{2} t} + c_2 e^{2t}$$

and the book's form

$$y(t)=c_1e^{r_1(t-t_0)}+c_2e^{r_2(t-t_0)}$$

are equivalent. $r_1=\frac{-5}2$ and $r_2=2$, which is obvious. Less obvious is that the $c$ constants take care of the book's $t_0$. In particular, your $c_1$ equals the book's $c_1e^{-rt_0}$ and your $c_2$ equals the book's $c_2e^{-rt_0}$, so there is no real difference between the two forms. Work with whichever one is more convenient for you.

If you use the book's form, you get the equations in $c_1$ and $c_2$ that you showed. If you used your original forms, you would have gotten for the first equation

$$c_1e^{\frac{-5}2}+c_2e^2=5$$

You see the book's form is indeed easier, so the way you were doing it is both right and the easier way.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

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