# D.E.: Boundary Value Problem

1. Feb 25, 2013

### Jeff12341234

I'm not sure if my answer is correct. Did I make a mistake somewhere? I'm not sure the ± needs to be there.

2. Feb 25, 2013

### HallsofIvy

Staff Emeritus
How in the world did you get a quadratic equation out of this? $y(2)= (C_1+ 6C_2)e^6= e^6$ and $y'(1)= (3C_1+ 4C_2)e^3= e^3$. The derivative is $y'= 3C_1e^{3x}+ C_2e^{3x}+ 3C_2xe^{3x}= ([3C_1+ C_2]+ 3C_2x)e^x$. It does not involve "$C_1C_2$"!

You have $C_1+ 6C_2= 1$ and $3C_1+ 4C_2= 1$, two linear equations.

3. Feb 25, 2013

### Jeff12341234

c1 is represented by c, c2 is represented by d

That's y'

I did make an error by leaving out the + sign between c1 and c2 for y'

That makes c1 = -1 and c2 = 1

Last edited: Feb 25, 2013