Determine the longest interval in which the given initial value

verskalksv 2021-11-15 Answered
Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the slotuion.
\(\displaystyle{\left({x}-{2}\right)}{y}{''}+{y}'+{\left({x}-{2}\right)}{\left({\tan{{x}}}\right)}{y}={0}\)
\(\displaystyle{y}{\left({3}\right)}={1}\)
\(\displaystyle{y}'{\left({3}\right)}={6}\)

Expert Community at Your Service

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

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Marian Tucker
Answered 2021-11-16 Author has 1168 answers
The given IVP
\(\displaystyle{\left({x}-{2}\right)}{y}{''}+{y}'+{\left({x}-{2}\right)}{\left({\tan{{x}}}\right)}{y}={0}\)
\(\displaystyle\Rightarrow{y}{''}+{\frac{{{1}}}{{{x}-{2}}}}{y}'+{\left({\tan{{x}}}\right)}{y}={0}\)
And the initial condition. \(\displaystyle{y}{\left({3}\right)}={1},\ {y}'{\left({3}\right)}={6}\)
Here \(\displaystyle{P}{\left({x}\right)}={\frac{{{1}}}{{{x}-{2}}}}\)
\(\displaystyle{q}{\left({x}\right)}={\tan{{x}}}\)
\(\displaystyle{g{{\left({x}\right)}}}={0}\) constant function is continuous everywhere.
Solution: \(\displaystyle{\left({\left({2}{n}+{1}\right)}{\frac{{\pi}}{{{2}}}},{\left({2}{n}+{3}\right)}{\frac{{\pi}}{{{2}}}}\right)}\)
Have a similar question?
Ask An Expert
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 2021-11-11
Determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution. Do not attempt to find the solution.
\(\displaystyle{\left({x}+{3}\right)}{y}{''}+{x}{y}'+{\left({\ln}{\left|{x}\right|}\right)}{y}={0},\ {y}{\left({1}\right)}={0},\ {y}'{\left({1}\right)}={1}\)
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 2021-10-23
Find the general solution of the given differential equation. \(\displaystyle{y}”−{2}{y}'−{3}{y}={3}{e}^{{{2}{t}}}\)
asked 2021-01-10
Using the existence and uniqueness theorem for second order linear ordinary differential equations, find the largest interval in which the solution to the initial value is certain to exist.
\(t(t^2-4)y''-ty'+3t^2y=0, y(1)=1 y'(1)=3\)
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 2021-11-20
Show that \(\displaystyle{y}={x}^{{2}}{\sin{{\left({x}\right)}}}\) and \(\displaystyle{y}={0}\) are both solutions of
\(\displaystyle{x}^{{2}}{y}{''}-{4}{x}{y}'+{\left({x}^{{2}}+{6}\right)}{y}={0}\)
and that both satisfy the conditions \(\displaystyle{y}{\left({0}\right)}={0}\) and \(\displaystyle{y}'{\left({0}\right)}={0}\). Does this theorem contradict Theorem A? If not, why not?
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?

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question
...