# Find a solution of the second-order IVP

• December 2nd 2009, 09:12 PM
DHS1
Find a solution of the second-order IVP question (First chapter in my Diff Eq's book)
Hey all. I have this problem:

$x=c_{1}e^x+c_2e^{-x}$

I am supposed to use this to find out what $c_1$ and $c_2$ are:

$y(0)=1$
$y'(0)=2$

What happens when I use the first equation is I get
$1=c_{1}e^0+c_2e^{-0}$
$1=c_{1}(1)+c_2(1)$
$1=c_{1}+c_2$

leaving me know way to solve for them. The answer in the back of the book says:

$y=\frac{3}{2}e^x-\frac{1}{2}e^{-x}$

Can anyone show me what I should do? If this one is answered, I will be able to complete all the other problems. Because I'm getting stuck on the same thing, having 1 equation and 2 variables $c_1$ and $c_2$

Thanks,
DHS1
• December 2nd 2009, 09:31 PM
Drexel28
Quote:

Originally Posted by DHS1
Hey all. I have this problem:

$x=c_{1}e^x+c_2e^{-x}$

I am supposed to use this to find out what $c_1$ and $c_2$ are:

$y(0)=1$
$y'(0)=2$

What happens when I use the first equation is I get
$1=c_{1}e^0+c_2e^{-0}$
$1=c_{1}(1)+c_2(1)$
$1=c_{1}+c_2$

leaving me know way to solve for them. The answer in the back of the book says:

$y=\frac{3}{2}e^x-\frac{1}{2}e^{-x}$

Can anyone show me what I should do? If this one is answered, I will be able to complete all the other problems. Because I'm getting stuck on the same thing, having 1 equation and 2 variables $c_1$ and $c_2$

Thanks,
DHS1

I assume you meant $y=C_1e^x+C_2e^{-x}$. Just plug and calculate $y(0)=C_1e^0+C_2e^{-0}=C_1+C_2=1$ and $y'(0)=C_1e^{0}-C_2e^{-0}=C_1-C_2=2$
• December 2nd 2009, 09:40 PM
Calculus26
See attachment