Consider, ay''+by'+cy=0 and a\ne0 Which of the following statements are

3kofbe 2021-11-23 Answered
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.

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Expert Answer

Thomas Conway
Answered 2021-11-24 Author has 405 answers
The general solution to this equation is,
\(\displaystyle{y}={C}_{{1}}{e}^{{\lambda_{{1}}{x}}}+{C}_{{2}}{e}^{{\lambda_{{2}}{x}}}\)
where, \(\displaystyle\lambda_{{1}}\) and \(\displaystyle\lambda_{{2}}\) are roots of the equation \(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}={0}\)
1) if we know \(\displaystyle{y}{\left({0}\right)}\) and \(\displaystyle{y}'{\left({0}\right)}\), then we can obtain two linear equations in \(\displaystyle{C}_{{1}}\) and \(\displaystyle{C}_{{2}}\), giving a unique solution.
2) As this function is exponential, we can say it is differentiable everywhere.
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Witheyesse47
Answered 2021-11-25 Author has 264 answers
the reason the (1) and (2) are correct is the nonsingular \(\displaystyle{\left({a}\ne{0}\right)}\)) linear equations have the uniqueness and existence property. all \(\displaystyle{y},{y}',{y}{''}\) stay bounded in the finite part of the domain. it always has two linearly independent solutions \(\displaystyle{y}_{{1}},{y}_{{2}}\) with \(\displaystyle{y}_{{1}}{\left({0}\right)}={1},\ {y}_{{1}}'{\left({0}\right)}={0},\ {y}_{{2}}{\left({0}\right)}={0},\ {y}_{{2}}'{\left({0}\right)}={1}\) so that they can take care of any initial conditions. like the you have \(\displaystyle{y}{\left({0}\right)}=\pi,\ {y}'{\left({0}\right)}=\sqrt{{\pi}}\)
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