Boundary Value Problem for y'' -2y' + 2y = 0: Help and Solution Guide

[bobmerhebi]

Homework Statement

solve the BVP: y'' -2y' + 2y = 0 ... (1)
reduction
y(0) = 1
y(pi) = 1

Homework Equations

y= e^mx

The Attempt at a Solution

substitution y=e^mx & its derivatives in (1) we get:

m² -2m +2 = 0 = (1+i)(1-i)

then y = e(c₁ sinx + c₂ cosx)

using the conditions given we have:

y(0) = 1 = c₂ cosx implies: c₂ = 1/e
&
y(pi) = 1 = -ec₂ implies: c₂ = -1/e

we thus have that c₂ = 1/e = -1/e which is impossible & thus the BVP has apparently no solution.

Is my solution correct?
Note that I am taking a course in Ordinary D.E.'s

thx in advance

 

[Lyuokdea]

² -2m +2 = 0 = (1+i)(1-i)

I assume you mean that this factors as m={(1+i),(1-i)}, and not that 0=(1+i)(1-i), which doesn't make sense.

now you have solutions for m, which you can plug back into your assumption that y=e^{mx}, i believe what you're missing is that you are factoring that solution for m incorrectly in order to obtain the y's

~Lyuokdea

 

[bobmerhebi]

actually yes. what i meant is that m₁ = 1+i & m₂ = 1-i

moreover, the general solution when the roots are complex is of the form:

y = e^\alpha { c₁ sin (\beta x) + c₂ cos(\beta x) }

with \alpha = 1 & \beta = 1 in this exercise.

so the solution becomes as previously written.

 

[HallsofIvy]

Yes, that is exactly what happens. This is an important difference between "initial value problems" and "boundary value problems". We have an "existence and uniqueness theorem" that says that as long as the differential equation is nice there exist a unique solution no matter what the initial conditions are.

With boundary value problems it is not so simple. For very simple differential equations you can have boundary conditions for which there is no solution or boundary value conditions for which there are an infinite number of solutions.

A simpler example is y"+ y= 0 with y(0)= 0, y(pi)= 1 which has no solution or y"+ y= 0 with y(0)= 0, y(pi)= 0 which has an infinite number of solutions.

Think of it as computing the trajectory of a bullet fired from a gun (which would give a fairly simple differential equation). Setting the gun at a specific point and aiming it in a given direction is an example of an "initial value" problem. The bullet will follow a single trajectory whether we can calculate it or not. Setting the gun at a specific point and requiring that the bullet hit a specific point on the target is an example of a "boundary value" problem. It may happen that the bullet can't hit that point (the target is beyond its range) or that there may be two different trajectories that will hit it (one above 45 degrees and the other below).

 

[bobmerhebi]

so this BVP i have with those conditions has no solution.