I need to solve the following differential equation x^2y''+(ax-b)y'-ay=0 with a,b>0,x\geq0 and

veksetz 2021-12-17 Answered
I need to solve the following differential equation
\(\displaystyle{x}^{{2}}{y}{''}+{\left({a}{x}-{b}\right)}{y}'-{a}{y}={0}\)
with \(\displaystyle{a},{b}{>}{0},{x}\geq{0}\) and \(\displaystyle{y}{\left({0}\right)}={0}\). The power series method will fail since there is a singularity at x=0, while the form of the equation does not conform with the Frobenius method.

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

Expert Answer

Kindlein6h
Answered 2021-12-18 Author has 3362 answers

\(\displaystyle{x}^{{2}}{y}{''}+{\left({a}{x}-{b}\right)}{y}'-{a}{y}={0}\)
Start by removing the leading behavior for small x. Near x=0 the leading terms in the equation are
\(\displaystyle-{b}{y}'-{a}{y}\approx{0}\)
(this is valid as long as in the end, as \(\displaystyle{x}\to{0},{y}{''}\) does not grow as fast as \(\displaystyle{y}\frac{'}{{x}^{{2}}}\) or \(\displaystyle\frac{{y}}{{x}^{{2}}}\), does not grow as fast \(\displaystyle{\frac{{y}}{{x}}}\)).
The solution to the approximate equation is \(\displaystyle{y}={e}^{{-{\frac{{a}}{{b}}}{x}}}\) which motivates the substitution
\(\displaystyle{y}={u}{\left({x}\right)}{e}^{{-{\frac{{a}}{{b}}}{x}}}\)
\(\displaystyle{y}'={u}'{e}^{{-{\frac{{a}}{{b}}}{x}}}-{\frac{{a}}{{b}}}{u}{e}^{{-{\frac{{a}}{{b}}}{x}}}\)
\(\displaystyle{y}{''}={u}{''}{e}^{{-{\frac{{{a}}}{{{b}}}}{x}}}-{2}{\frac{{{a}{b}}}{{u}}}'{e}^{{-{\frac{{a}}{{b}}}{x}}}+{\frac{{{a}^{{2}}}}{{{b}^{{2}}}}}{u}{e}^{{-{\frac{{a}}{{b}}}{x}}}\)
and the differential equation becomes
\(\displaystyle{x}^{{2}}{u}{''}-{\left({2}{\frac{{{a}}}{{{b}}}}{x}^{{2}}-{a}{x}+{b}\right)}{u}'+{\left({\frac{{{a}^{{2}}}}{{{b}^{{2}}}}}{x}^{{2}}-{\frac{{{a}^{{2}}}}{{{b}}}}{x}+{a}-{a}\right)}{u}={0}\)
with u(0)=0
Now if we let \(\displaystyle{u}{\left({x}\right)}={\sum_{{0}}^{\infty}}{u}_{{k}}{x}^{{k}}\)
\(\displaystyle{\sum_{{0}}^{{\infty}}}{k}{\left({k}-{1}\right)}{u}_{{k}}{x}^{{k}}-{2}{\frac{{a}}{{b}}}{\sum_{{0}}^{\infty}}{k}{u}_{{k}}{x}^{{{k}+{1}}}+{a}{\sum_{{0}}^{{\infty}}}{k}{u}_{{k}}{x}^{{k}}\)-\(b\sum ku_kx^{k-1}+\frac{a^2}{b^2}\sum_0^\infty u_kx^{k+2}-\frac{a^2}{b}\sum_0^\infty u_kx^{k+1}=0\)
And in each term we can substitute an offset index to always get \(\displaystyle{x}^{{k}}\) this gives
\(\displaystyle{\left[{k}{\left({k}-{1}\right)}+{a}{k}\right)}{u}_{{k}}-{\left[{2}{\frac{{{a}}}{{{b}}}}{\left({k}-{1}\right)}+{\frac{{{a}^{{2}}}}{{{b}}}}\right]}{u}_{{{k}-{1}}}-{b}{\left({k}+{1}\right)}{u}_{{{k}+{1}}}+{\frac{{{a}^{{2}}}}{{{b}^{{2}}}}}{u}_{{{k}-{2}}}={0}\)
and this is to be solved with \(\displaystyle{u}_{{-{1}}}={u}_{{0}}={0}\)
there is on free parameter \(\displaystyle{u}_{{1}}\)
\(\displaystyle{u}_{{2}}={\frac{{{a}{u}_{{1}}}}{{{b}}}}\)
and so forth. So the solutions are this series, which is well behaved near zero, times \(\displaystyle{e}^{{-{\frac{{a}}{{b}}}{x}}}\)
All of the coefficients are proportional to \(\displaystyle{u}_{{1}}\). The overall solution has one free parameter and is then
\(\displaystyle{y}={u}_{{1}}{e}^{{-{a}\frac{{x}}{{b}}}}{\left[{x}+{\frac{{a}}{{b}}}{x}^{{2}}+{\frac{{23}}{{\frac{{{2}^{{2}}}}{{{b}^{{2}}}}}}}{x}^{{3}}+{\frac{{{5}{a}^{{3}}+{3}{a}^{{2}}}}{{{12}{b}^{{2}}}}}{x}^{{4}}+{\frac{{{17}{a}^{{4}}+{39}{a}^{{3}}+{36}}}{{{60}{b}^{{4}}}}}{x}^{{5}}+\ldots\right]}\)
\(\displaystyle={u}_{{1}}{\left[{x}+{\frac{{{a}^{{2}}}}{{{6}{b}^{{2}}}}}{x}^{{3}}+{\frac{{{a}^{{3}}+{3}{a}^{{2}}}}{{{12}{b}^{{3}}}}}{x}^{{4}}+{\frac{{{3}{a}^{{4}}+{16}{a}^{{3}}+{24}{a}^{{2}}}}{{{60}{b}^{{4}}}}}{x}^{{5}}+\ldots\right]}\)
As to whether this series is convergent for all positive x, or has a finite radius of convergence, that depends on the values of a and b.

Not exactly what you’re looking for?
Ask My Question
0
 
Gerald Lopez
Answered 2021-12-19 Author has 1147 answers
It has a trivial solution \(\displaystyle{y}={a}{x}-{b}\)
Let \(\displaystyle{y}={\left({a}{x}-{b}\right)}{u}\)
Then \(\displaystyle{y}'={\left({a}{x}-{b}\right)}{u}'+{a}{u}\)
\(\displaystyle{y}{''}={\left({a}{x}-{b}\right)}{u}{''}+{a}{u}'+{a}{u}'={\left({a}{x}-{b}\right)}{u}{''}+{2}{a}{u}'\)
\(\displaystyle{x}^{{2}}{\left({\left({a}{x}-{b}\right)}{u}{''}+{2}{a}{u}'\right)}+{\left({a}{x}-{b}\right)}{\left({\left({a}{x}-{b}\right)}{u}'+{a}{u}\right)}-{a}{\left({a}{x}-{b}\right)}{u}={0}\)
\(\displaystyle{x}^{{2}}{\left({a}{x}-{b}\right)}{u}{''}+{2}{a}{x}^{{2}}{u}'+{\left({a}{x}-{b}\right)}^{{2}}{u}'+{a}{\left({a}{x}-{b}\right)}{u}-{a}{\left({a}{x}-{b}\right)}{u}={0}\)
\(\displaystyle{x}^{{2}}{\left({a}{x}-{b}\right)}{u}{''}=-{\left({2}{a}{x}^{{2}}+{\left({a}{x}-{b}\right)}^{{2}}\right)}{u}'\)
\(\displaystyle{\frac{{{u}{''}}}{{{u}}}}=-{\frac{{{2}{a}}}{{{a}{x}-{b}}}}-{\frac{{{a}}}{{{x}}}}+{\frac{{{b}}}{{{x}^{{2}}}}}\)
\(\displaystyle{\ln{{u}}}'=-{2}{\ln{{\left({a}{x}-{b}\right)}}}-{a}{\ln{{x}}}-{\frac{{{b}}}{{{x}}}}+{c}\)
\(\displaystyle{u}'={\frac{{{C}{e}^{{-{\frac{{{b}}}{{{x}}}}}}}}{{{x}^{{a}}{\left({a}{x}-{b}\right)}^{{2}}}}}\)
\(\displaystyle{u}={C}_{{1}}+{C}_{{2}}\int^{{x}}{\frac{{{e}^{{-{\frac{{{b}}}{{{x}}}}}}}}{{{x}^{{a}}{\left({a}{x}-{b}\right)}^{{2}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle{y}={C}_{{1}}{\left({a}{x}-{b}\right)}+{C}_{{2}}{\left({a}{x}-{b}\right)}\)
\(\displaystyle\int^{{x}}{\frac{{{e}^{{-{\frac{{{b}}}{{{x}}}}}}}}{{{x}^{{a}}{\left({a}{x}-{b}\right)}^{{2}}}}}{\left.{d}{x}\right.}\)
0
RizerMix
Answered 2021-12-29 Author has 9350 answers

With Laplace transforms this equation will be
\(s^2F''(s)+(4-a-b)sF'(s)+2(1-a)F(s)=0 \)
which is Euler's equation.

0

Expert Community at Your Service

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

Relevant Questions

asked 2021-09-17
Solve each of the differential equations in Table 1. Include the characterestic polynomial and its roots with your answer.
\(\displaystyle{y}{''}-{3}{y}'+{2}{y}={0}\)
asked 2021-11-23
I have the following differential equation:
\(\displaystyle{y}{''}+{y}={\cos{{\left({t}\right)}}}{\cos{{\left({2}{t}\right)}}}\)
Maybe something can be done to \(\displaystyle{\cos{{\left({t}\right)}}}{\cos{{\left({2}{t}\right)}}}\) to make it easier to solve. Any ideas?
asked 2021-10-23
Find the general solution of the given differential equation. \(\displaystyle{y}”−{2}{y}'−{3}{y}={3}{e}^{{{2}{t}}}\)
asked 2021-11-23
Consider, \(\displaystyle{a}{y}{''}+{b}{y}'+{c}{y}={0}\) and \(\displaystyle{a}\ne{0}\) Which of the following statements are always true?
1. A unique solution exists satisfying the initial conditions \(\displaystyle{y}{\left({0}\right)}=\pi,\ {y}'{\left({0}\right)}=\sqrt{{\pi}}\)
2. Every solution is differentiable on the interval \(\displaystyle{\left(-\infty,\infty\right)}\)
3. If \(\displaystyle{y}_{{1}}\) and \(\displaystyle{y}_{{2}}\) are any two linearly independent solutions, then \(\displaystyle{y}={C}_{{1}}{y}_{{1}}+{C}_{{2}}{y}_{{2}}\) is a general solution of the equation.
asked 2021-11-21
I need to find a second order linear homogeneous equation with constant coefficients that has the given function as a solution
Queston a) \(\displaystyle{x}{e}^{{-{3}{x}}}\)
Question b) \(\displaystyle{e}^{{{3}{x}}}{\sin{{x}}}\)
asked 2021-11-19
We have the following differential equation
\(\displaystyle\cup{''}={1}+{\left({u}'\right)}^{{2}}\)
i found that the general solution of this equation is
\(\displaystyle{u}={d}{\text{cosh}{{\left({\left({x}-\frac{{b}}{{d}}\right)}\right.}}}\)
where b and d are constats
Please how we found this general solution?
asked 2021-11-19
I need some hints for solving \(\displaystyle{y}{y}{''}-{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\)
I noticed that the left hand side is close to \(\displaystyle{\left({y}{y}'\right)}'\):
\(\displaystyle{y}{y}{''}-{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\Rightarrow{y}{y}{''}+{\left({y}'\right)}^{{2}}-{2}{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\Rightarrow{\left({y}{y}'\right)}'-{2}{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\)
But I don't know how to continue expressing the terms as derivatives of some functions.
...