How can the following second-order linear equation be converted into

chanyingsauu7 2021-11-21 Answered
How can the following second-order linear equation be converted into a first-order linear equation?
This is our second-order equation:
\(\displaystyle{y}{''}-{2}{y}'+{2}{y}={e}^{{{2}{t}}}{\sin{{t}}}\)

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

soniarus7x
Answered 2021-11-22 Author has 1746 answers
Note that if \(\displaystyle\alpha\) and \(\displaystyle\beta\) are the roots of \(\displaystyle{x}^{{2}}+{a}{x}+{b}={0}\), we have \(\displaystyle\alpha=-{\left(\alpha+\beta\right)}\) and \(\displaystyle{b}=\alpha\beta\).
The equation \(\displaystyle{y}{''}+{a}{y}'+{b}{y}={f{{\left({t}\right)}}}={y}{''}-{\left(\alpha+\beta\right)}{y}'+\alpha\beta{y}\)
We now rewrite the left-hand side as \(\displaystyle{y}{''}-\alpha{y}'-\beta{y}'+\alpha\beta{y}={\left({y}{''}-\alpha{y}'\right)}-\beta{\left({y}'-\alpha{y}\right)}={f{{\left({t}\right)}}}\) and substitute \(\displaystyle{z}={y}'-\alpha{y}\) so that \(\displaystyle{z}'={y}{''}-\alpha{y}'\) and finally the transformed equation becomes
\(\displaystyle{z}'-\beta{z}={f{{\left({t}\right)}}}\)
Once \(\displaystyle{z}={g{{\left({t}\right)}}}\) is known, y is obtained from \(\displaystyle{y}'-\alpha{y}={g{{\left({t}\right)}}}\)
So one second order equation can be solved by solving two first order equations.
Have a similar question?
Ask An Expert
0
 
Huses1969
Answered 2021-11-23 Author has 338 answers
Look at the underlying homogeneous equation \(\displaystyle{y}{''}-{2}{y}'+{2}={0}\); it's characteristic polynomial equation is the quadratic \(\displaystyle\lambda^{{2}}-{2}\lambda+{2}={0}\); the discriminant of this quadratic is \(\displaystyle{\left(-{2}\right)}^{{2}}-{4}{\left({2}\right)}={4}-{8}=-{4}{<}{0}\)</span>, so the equation has a pair of complex confugate roots, they are in fact \(\displaystyle{1}\pm{i}\). Thus the (real) dimension of the solution space is 2, and it can't be reduced. Therefore, since the equation is intrinsically possessed of two degrees of freedom, typically manifested as \(\displaystyle{y}{\left({0}\right)}\) and \(\displaystyle{y}'{\left({0}\right)}\), the only way to lower the order is by expressing the original equation as a first order system on the two-dimensional space \(\displaystyle{\mathbb{{{R}}}}^{{2}}\). Thus we set
\(\displaystyle{z}{\left({t}\right)}={y}'{\left({t}\right)}\)
so that
\(\displaystyle{z}'{\left({t}\right)}={2}{z}{\left({t}\right)}-{2}{y}{\left({t}\right)}+{e}^{{{2}{t}}}{\sin{{t}}}\)
In vector-matrix form this may be written as
\[\left(\begin{array}{c}y(t)\\z(t)\end{array}\right)'=\begin{bmatrix}0 & 1 \\-2 & 2 \end{bmatrix}\left(\begin{array}{c}y(t)\\ z(t)\end{array}\right)+\left(\begin{array}{c}0\\ e^{2t}\sin t\end{array}\right)\]
A first order system equivalent in all ways to the original second order ODE
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 2021-11-21
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
I'm trying to find particular solution of the second-order linear equation but I can't find \(\displaystyle{y}_{{1}}\) and \(\displaystyle{y}_{{2}}\) according to \(\displaystyle{y}={c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}\)
\(\displaystyle{x}^{{2}}{y}{''}-{2}{x}{y}'+{2}{y}={0},\ {y}{\left({1}\right)}={3},\ {y}'{\left({1}\right)}={1}\)
If r is used, \(\displaystyle{x}^{{2}}{r}^{{2}}-{2}{x}{r}=-{2}\) then \(\displaystyle{x}{r}{\left({x}{r}-{2}\right)}=-{2}\), I can't go on from there to find \(\displaystyle{y}_{{1}}\) and \(\displaystyle{y}_{{2}}\)
asked 2021-11-23
Solve th second order linear equations:
\(\displaystyle{y}{''}+{4}{y}={\cos{{2}}}{x}\)
asked 2021-11-22

Find second order linear homogeneous ODE with constant coefficients if its fundamental set of solutions is \(\{e^{3t},te^{3t}\}\)

asked 2021-11-20
Is there any known method to solve such second order non-linear differential equation?
\(\displaystyle{y}{''}_{{n}}-{n}{x}{\frac{{{1}}}{{\sqrt{{{y}_{{n}}}}}}}={0}\)
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 2021-11-19
Consider the following linear difference equation
\(\displaystyle{f}_{{k}}={1}+{\frac{{{1}}}{{{2}}}}{f}_{{{k}+{1}}}+{\frac{{{1}}}{{{2}}}}{f}_{{{k}-{1}}},\ {1}\le{k}\le{n}-{1}\)
with \(\displaystyle{f}_{{0}}={f}_{{n}}={0}\). How do i find 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
...