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Linear homogenous ODEs with constant coefficients 
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#1
Nov1212, 01:39 PM

P: 103

Given the ODE of the form:
y''(x) + A*y'(x) + B*y(x) = 0 If we choose a solution such that y(x) = e[itex]^{mx}[/itex] and plug it into the original ODE, the ODE becomes: (m[itex]^{2}[/itex] + A*m + B)e[itex]^{mx}[/itex] = 0 If we solve for the roots of the characteristic equation such that m = r[itex]_{1}[/itex], r[itex]_{2}[/itex] (root 1 and root 2, respectively) The solution to the ODE would have the form: y(x) = c*e[itex]^{r_{1}*x}[/itex] + d*e[itex]^{r_{2}*x}[/itex], where c and d are constants My question is, why are the constants where they are in the solution? In other words, why are they multiplying y[itex]_{1}[/itex] & y[itex]_{2}[/itex], where y[itex]_{1}[/itex] = e[itex]^{r_{1}*x}[/itex] and y[itex]_{2}[/itex] = e[itex]^{r_{2}*x}[/itex] ? Why are the constants not just added to y(x), such that the solution to the ODE would be as follows, where k is a constant. y(x) = e[itex]^{r_{1}*x}[/itex] + e[itex]^{r_{2}*x}[/itex] + k This question mainly is for second order and higher differential equations. I understand how it works for first order linear homogenous DEs because the constant is simply the constant of integration, but I am having trouble understanding how it applies to higher orders. 


#2
Nov1212, 07:27 PM

HW Helper
P: 2,263

That is because the differential equation is linear. That is if L[y]=0 is a linear differential equation and u and v are any two solutions so that L[u]=L[v]=0 then L[a u+a v]=0.



#3
Nov1312, 07:58 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,338

The fundamental theorem for such equations is:
"The set of all solutions to a linear homogeneous differential equation of order n form a vector space of dimension n" That means that if we can find a set of n independent solutions, a basis for that vector space of solutions, any solution can be written as a linear combination of those solutions. And a "linear combination" means a sum of the functions multiplied by constants. 


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