Second Order Differential Equations ay''+by'+cy=0, without complex numbers

sexe4yo 2022-05-02 Answered
Second Order Differential Equations ay+by+cy=0, without complex numbers
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Addison Zamora
Answered 2022-05-03 Author has 10 answers
Step 1
Let us show that there is a unique solution, given that
g(0)=A,
g(0)=B.
Clearly Acosx+Bsinx is a solution with these initial conditions, so look at h(x):g(x)Acosx+Bsinx; this satisfies h+h=0, h(0)=0, h(0)=0. We want to show that h(x)=0 for all x.
Multiplying h+h=0 by h we see that hh+hh=0 for all x. Note that hh+hh=(12h2+12h2). Therefore, by the Mean Value Theorem,
12h(x)2+12h(x)2=(x0)(12h(θx)2+12h(θx)2=0.
Now a real sum of squares can only be zero if each summand is zero. So we have, as required, h(x)=0 for all x.
[For those of an applied bent, we are just doing the classical thing of conserving the total energy.]
Not exactly what you’re looking for?
Ask My Question
Makayla Santiago
Answered 2022-05-04 Author has 19 answers
Step 1
You can make the educated guess that the solution will be of the form
f(x)=ekxg(x).
Then we get
a[ekxg(x))+b(ekxg(x))+cekxg(x)
=a[ekxg(x)+2kekxg(x)+k2ekxg(x)]+b[ekxg(x)+kekxg(x)]+cekxg(x)
=aekxg(x)+[2ak+b]ekxg(x)+[ak2+bk+c]ekxg(x)
=0
Since ekx is never 0, this reduces to
ag+(2ak+b)g+(ak2+bk+c)g=0.
We are free to choose k as we wish, because if ekxg(x) solves the ODE, then so does ek~xe(kk~)xg(x), and we simply get g(x)~=e(kk~)g(x) as the solution for g instead. So we choose k as simple as possible: such that 2ak+b=0, that is, k=b2a. Then the equation reads (after multiplying by 4a)
4a2g(b24ac)g=0.
The solution to this, given that b24ac<0, is known: it's of the form g(x)=Asin(ωx)+Bcos(ωx), where ω=cb24a.
So in total we get f(x)=ekx[Asin(ωx)+Bcos(ωx)] with k and ω as specified.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more