# initial value problem

• Dec 29th 2010, 05:26 AM
immortality
initial value problem
Given that y''=y'+2y and that y(0)=0 and y(1)=1 Deduce that one of the initial value problems has a trivial solution.

Please tell me how can I get the trivial answer. If we set y(0)=0 and y'(0)=0 we get the correct answer, but the answer is saying that y(0)=0 and y'(0)=1 is also trivial(!) which doesn't make sense to me.
• Dec 29th 2010, 05:42 AM
hatsoff
An IVP, if I am not mistaken, refers to an equation together with a SINGLE value. What you have there looks more like a boundary problem. In any case, it has two values, so it's not an IVP.

Perhaps they mean to give you two IVPs with the same equation... ? In that case y(0)=0 is obviously the one with the trivial solution.
• Dec 31st 2010, 04:42 PM
dwsmith
Quote:

Originally Posted by immortality
Given that y''=y'+2y and that y(0)=0 and y(1)=1 Deduce that one of the initial value problems has a trivial solution.

Please tell me how can I get the trivial answer. If we set y(0)=0 and y'(0)=0 we get the correct answer, but the answer is saying that y(0)=0 and y'(0)=1 is also trivial(!) which doesn't make sense to me.

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

$\displaystyle y'(x)=\frac{(2e^{3x}+1)e^{1-x}}{e^3-1}$

$\displaystyle y'(0)=\frac{(2e^{0}+1)e^{1}}{e^3-1}=\frac{3e}{e^3-1}\neq 1$
• Jan 1st 2011, 02:47 AM
HallsofIvy
Quote:

Originally Posted by immortality
Given that y''=y'+2y and that y(0)=0 and y(1)=1 Deduce that one of the initial value problems has a trivial solution.

You are given a boundary value problem. I don't know what "one of the initial value problems" means. One of what initial value problems? Certainly one possible initial value problem for this differential equation is "y''= y'+ 2y with y(0)= 0, y'(0)= 0". That uses the same equation and one of the given boundary values so perhaps that is what is meant.

Quote:

Please tell me how can I get the trivial answer. If we set y(0)=0 and y'(0)=0 we get the correct answer, but the answer is saying that y(0)=0 and y'(0)=1 is also trivial(!) which doesn't make sense to me.
• Jan 1st 2011, 02:49 AM
HallsofIvy
Quote:

Originally Posted by hatsoff
An IVP, if I am not mistaken, refers to an equation together with a SINGLE value. What you have there looks more like a boundary problem. In any case, it has two values, so it's not an IVP.

An "initial value problem" for a second order equation would give a value of y and its derivative at a single value of the independent variable.

Quote:

Perhaps they mean to give you two IVPs with the same equation... ? In that case y(0)=0 is obviously the one with the trivial solution.