# Find a solution of the second-order IVP

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• Dec 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:

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

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

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

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

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

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

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 $\displaystyle c_1$ and $\displaystyle c_2$

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

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

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

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

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

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

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

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

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 $\displaystyle c_1$ and $\displaystyle c_2$

Thanks,
DHS1

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