why are the solutions to System of First Order DE's of the form x= Σe^(rt)?

[kochibacha]

In case of homogeneous, i have read more than 4 System of First Order Differential Equations books and most of the method
of finding the solutions are

"assume the solutions of the form x = Σe^rt,
"seek the solutions of the form x = Σe^rt",
"the idea is to find solutions of the form x = Σe^rt
etc.

could you explain in very detail why of the form x = Σe^rt



there are about 4 theories which explain the general solution of the system will be in the form c₁x⁽¹⁾+c₂x⁽¹⁾+...c_nx⁽ⁿ⁾ and they are unique
but why are x must be in the form x = Σe^rt?

same goes to second order linear equations are they just picking the y=e^rt, after some experiment they found that sum is also the solutions and they construct the method to solve second order DE's from some random solutions?. I know there would be some clue about how the solutions behave but come on there should be more generalized explanation than just random some solutions.

 

[JohnZ]

In case of homogeneous, i have read more than 4 System of First Order Differential Equations books and most of the method
of finding the solutions are

"assume the solutions of the form x=ξe^rt",
"seek the solutions of the form x=ξe^rt",
"the idea is to find solutions of the form x=ξe^rt"
etc.

could you explain in very detail why of the form x=ξe^rt



there are about 4 theories which explain the general solution of the system will be in the form c1x(1)+c2x(1)+...cnx(n) and they are unique
but why are x must be in the form x=ξe^rt?

same goes to second order linear equations are they just picking the y=e^rt, after some experiment they found that sum is also the solutions and they construct the method to solve second order DE's from some randomed solutions?. I know there would be some clue about how the solutions behave but come on there should be more generalized explanation than just random some solutions.

DE should be homogeneous and constant coefficient, then form of solution is e^rt.
suppose DE is dx/dt + a*x = 0.
plug x=e^rt in, get r*e^rt + a*e^rt = (r+a)*e^rt = 0, since e^rt can never be 0,
r+a must be 0 to make equation equals 0.
thus r=-a. we find a solution x=e^-at.
another theorem says all the solutions should be the form c*x(t), x(t) is a solution of DE.
now we know x=e^-at is a solution, then DE's all solutions can be expressed as c*e^-at

 

[HallsofIvy]

Notice that the books are not saying that the solutions are of that form. They are suggesting that you try solutions of that form. It should be no surprise that they work because the differential equations have been set up to give solutions of that form. Later you will see examples where the solutions are NOT of that form but the information you gain by trying a solution of that form will lead you to a solution.

 

[JohnZ]

Later you will see examples where the solutions are NOT of that form but the information you gain by trying a solution of that form will lead you to a solution.

could you give an example?

 

[Deleted member 4993]

could you give an example?

\(\displaystyle \dfrac{dy}{dx} \ = \ x \)

 

[kochibacha]

Notice that the books are not saying that the solutions are of that form. They are suggesting that you try solutions of that form. It should be no surprise that they work because the differential equations have been set up to give solutions of that form. Later you will see examples where the solutions are NOT of that form but the information you gain by trying a solution of that form will lead you to a solution.

so the idea is try solution of that form and then work up some theories to prove whether it's unique or there are possibility of the solutions right?. Anyways, i found some of the examples u have mentioned and again they pick out some form of te^rt believing it's one of the solution other than e^rt i think there are a theory or method behind this what if i have a very complicated one which one of the solution might be te^er^rt^3t^ert and there are other complicated solutions to this system as well. How can you pick these solutions? or this system could never be solve unless numerical method is employed?.

one more question: from this theory which states that " if vector function x1, x2 ... xn is solution of the system then x1+x2+..xn is also a solution of the system" i wonder if i found x1+x2 =x3 first then i found x1 and x2 afterward, from the theory x3+x1+x2 = 2(x1)+2(x2) also a solution, repeat this will give and infinite sum of the terms x1 and x2. How can this be explained?

 

[JohnZ]

dydx = x

this is not homogeneous DE.

 

[JohnZ]

so the idea is try solution of that form and then work up some theories to prove whether it's unique or there are possibility of the solutions right?. Anyways, i found some of the examples u have mentioned and again they pick out some form of te^rt believing it's one of the solution other than e^rt i think there are a theory or method behind this what if i have a very complicated one which one of the solution might be te^er^rt^3t^ert and there are other complicated solutions to this system as well. How can you pick these solutions? or this system could never be solve unless numerical method is employed?.

for second order linear homogeneous constant coefficient DE. the solutions can be:
1) y=c1*e^(r1*x) + c2*e^(r2*x)
2) y=c1*e^(r*x) + c2*x*e^(r*x)
That' all.

 

[Deleted member 4993]

this is not homogeneous DE.

The homogeneous part is : \(\displaystyle \dfrac{dy}{dx} \ = \ 0\)

 

[JohnZ]

The homogeneous part is : dy/dx = 0

what you are trying to say?

 

[HallsofIvy]

what you are trying to say?

The point was that general solution to dy/dx= x is, by simply integrating both sides, is \(\displaystyle y= x^2/2\) which is NOT an exponential. The general solution to dy/dx is y= C, a constant. Of course that can be written in the form \(\displaystyle y= Ce^{0t}\) so it is in that sense an exponential.

I had in mind things like the system \(\displaystyle \frac{dx}{dt}= y\), \(\displaystyle \frac{dy}{dt}= -x\) which has general solution \(\displaystyle x= C cos(t)+ D sin(t)\), \(\displaystyle y= D cos(t)- C sin(t)\) and \(\displaystyle \frac{dx}{dt}= y\), \(\displaystyle \frac{dy}{dt}= 2y- x\) which has general solution \(\displaystyle x= Ce^{t}+ Dte^{t}\), \(\displaystyle y= (C+ D)e^t+ Dte^t\).

 

[JohnZ]

The point was that general solution to dy/dx= x is, by simply integrating both sides, is y=x 2 /2 which is NOT an exponential. The general solution to dy/dx is y= C, a constant. Of course that can be written in the form y=Ce 0t so it is in that sense an exponential.

I had in mind things like the system dxdt =y , dydt =−x which has general solution x=Ccos(t)+Dsin(t) , y=Dcos(t)−Csin(t) and dxdt =y , dydt =2y−x which has general solution x=Ce t +Dte t , y=(C+D)e t +Dte t .

we are here to discuss homogeneous equation. but dy/dx = x is not homogeneous equation.

dx/dt =y , dy/dt =−x can be convert to 2nd order DE: x'' + x = 0. it's auxiliary equation r^2 + 1 = 0, roots are +/-i, solution x = c1*e^(i*t) + c2*e^(-i*t), and it is only another presentation for x=C*cos(t) + D*sin(t)