# Thread: differential equations help

1. ## differential equations help

if you guys can help for these 4 numbers it would be extremely nice

3. ## Re: differential equations help

Let's do this by easy steps:

Step 1: Take a differential equations class!

4. ## Re: differential equations help

I did all those using pencil and paper the other day. It took only a few minutes and required only that I follow the hints.

5. ## Re: differential equations help

Ahh, but you followed the hints. It is remarkable how many people simply cannot follow hints!

(You can see I am in a very sarcastic mood today.)

6. ## Re: differential equations help

Originally Posted by Musawwir
if you guys can help for these 4 numbers it would be extremely nice
I did 4a for you here

7. ## Re: differential equations help

5a) (The only one that does not have specific "hints") $y''- 4y'- 3y= xe^{|x|}$.

The "associated homogeneous equation" is $r^2- 4r- 3= (r- 3)(r- 1)= 0$ and has roots r= 1 and r= 3. The general solution to the associated homogeneous equation is $C_1e^x+ C_2e^{3x}$.

The "non-homogeneous part" is $xe^{|x|}$. To handle the "|x|", treat $x\ge 0$ and $x< 0$" separately.

If $x\ge 0$ then $xe^{|x|}= xe^x$. Try a "specific solution" of the form $y= (Ax+ B)e^x$. Then $y'= Ae^x+ (Ax+ B)e^x= (Ax+ A+ B)e^x$ and $y''= Ae^x+ (Ax+ A+ B)e^x= (Ax+ 2A+ B)e^x$.

So $y''- 4y'- 3y= xe^{x}$ becomes $(Ax+ 2A+ B)e^x- 4(Ax+ A+ B)e^x- 3(Ax+ B)e^x= (-6Ax- 2A- 5B)e^x= xe^x$. We must have -6A= 1 and -2A- 5B= 0. From the first equation, $A= -\frac{1}{6}$ and then the second equation becomes $\frac{1}{3}= 5B$ so $B= \frac{1}{15}$.

For $x\ge 0$ the general solution to the entire equation is $y(x)= C_1e^x+ C_2e^{3x}+ \left(-\frac{1}{6}x+ \frac{1}{15}\right)e^x$.

Now do the same, for x< 0, using $y= xe^{-x}$ as the "non-homogeneous" part, so looking for a "specific solution" of the form $(Ax+ B)e^{-x}$. Of course, the "undetermined coefficients", for this solution, might not be the same as the "$C_1$" and "$C_2$". Call then, say, "$D_1$" and "$D_2$". Use the fact that, since y must be twice differentiable (the equation involves y'') the function and its derivative must be continuous at x= 0 to determine "$D_1$" and "$D_2"$ in terms of "$C_1$" and "$C_2$".