I need to find a second order linear homogeneous equation

Edmund Adams 2021-11-21 Answered
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) \(\displaystyle{e}^{{{3}{x}}}{\sin{{x}}}\)

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Expert Answer

Hiroko Cabezas
Answered 2021-11-22 Author has 1747 answers
Question A Let the answer to question a be of the form
\(\displaystyle{a}{y}{''}+{b}{y}'+{c}{y}={0}\)
Then we want \(\displaystyle{a}{\left(-{3}\right)}^{{2}}-{3}{b}+{c}={9}{a}-{3}{b}+{c}={0}\) (1)
And more over, since it is a repeated root we need
\(\displaystyle-{\frac{{{b}}}{{{2}{a}}}}=-{3}\)
Substituting \(\displaystyle{b}={6}{a}\) into (1) we get
\(\displaystyle{9}{a}-{18}{a}+{c}={0}\Rightarrow={9}{a}\)
Thus if we let a be our free variable, all solutions to question a are of the form
\(\displaystyle{a}{y}{''}+{6}{a}{y}'+{9}{a}{y}={0}\)
Question B For question b, you need the characteristic polynomial to have the roots \(\displaystyle{3}\pm{i}\). Thus the characteristic polynomial must be of the form
\(\displaystyle{a}{\left({x}-{\left[{3}+{i}\right]}\right)}{\left({x}-{\left[{3}-{i}\right]}\right)}={a}{x}^{{2}}-{6}{a}{x}+{10}{a}\)
Thus solutions to question b are of the form
\(\displaystyle{a}{y}{''}-{6}{a}{y}'+{10}{a}{y}={0}\)
Note For your solution to question a you got \(\displaystyle{y}{''}-{6}{y}'+{9}{y}={0}\) but going back and checking you should get \(\displaystyle+{6}{y}'\) not \(\displaystyle-{6}{y}'\). Otherwise your solutions are of these forms with \(\displaystyle{a}={1}\) but a can be any real number. It is our free variable.
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Unpled
Answered 2021-11-23 Author has 1297 answers
Hints For the first one, \(\displaystyle\lambda_{{1}}=\lambda_{{2}}=-{3}\) are roots of the chair. polynomial , so
\(\displaystyle\chi{\left(\lambda\right)}={\left(\lambda-{\left(-{3}\right)}\right)}^{{2}}=\lambda^{{2}}+{6}\lambda+{9}\)
Which corresponds to the diff. eq.
\(\displaystyle{y}{''}+{6}{y}'+{9}{y}={0}\)
Which has, as already found, the general solution
\(\displaystyle{y}={c}_{{1}}{e}^{{-{3}{x}}}+{c}_{{2}}{e}^{{-{3}{x}}}\)
Now choose the values of \(\displaystyle{y}{\left({0}\right)}\) and \(\displaystyle{y}'{\left({0}\right)}\) in such a way that \(\displaystyle{c}_{{1}}={0}\) and \(\displaystyle{c}_{{2}}={1}\). With \(\displaystyle{y}{\left({0}\right)}\) we have
\(\displaystyle{y}{\left({0}\right)}={c}_{{1}}\)
So we can set \(\displaystyle{y}{\left({0}\right)}={0}\) to set \(\displaystyle{c}_{{1}}={0}\). For \(\displaystyle{y}'{\left({0}\right)}\) we have
\(\displaystyle{y}'{\left({x}\right)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({c}_{{2}}{x}{e}^{{-{3}{x}}}\right)}={c}_{{2}}{e}^{{-{3}{x}}}\pm{3}{x}{c}_{{2}}{e}^{{-{3}{x}}}={c}_{{2}}{e}^{{-{3}{x}}}{\left({1}-{3}{x}\right)}\)
Then
\(\displaystyle{y}'{\left({0}\right)}={c}_{{2}}\)
So we set \(\displaystyle{y}'{\left({0}\right)}={1}\) to get when we want.
For the second, you should know that if the solution \(\displaystyle\lambda\) to \(\displaystyle\chi{\left(\lambda\right)}\) looks like
\(\displaystyle{e}^{{{3}{x}}}{\sin{{\left({x}\right)}}}\)
Then deduce the characteristic polynomial from it, from which you can deduce the original differential equation.
Start by identifying \(\displaystyle\lambda_{{1}}={3}+{i}\), deduce the complex conjugate pair as \(\displaystyle{a}_{{2}}={3}-{i}\), now set
\(\displaystyle\chi{\left(\lambda\right)}={\left(\lambda-\lambda_{{1}}\right)}{\left(\lambda-\lambda_{{2}}\right)}\)
To get
\(\displaystyle\chi{\left(\lambda\right)}=\lambda^{{2}}-{6}\lambda+{10}\)
Which corresponds to \(\displaystyle{y}{''}-{6}{y}+{10}{y}={0}\)
Which has the general solution \(\displaystyle{y}{\left({x}\right)}={c}_{{1}}{e}^{{{3}{x}}}{\cos{{\left({x}\right)}}}+{c}_{{2}}{e}^{{{3}{x}}}{\sin{{\left({x}\right)}}}\)
Again, as above, choose y(0) and y'(0) in such a way that \(\displaystyle{c}_{{1}}={0}\) and \(\displaystyle{c}_{{2}}={1}\) to give the desired solution.
We see that \(\displaystyle{y}{\left({0}\right)}={c}_{{1}}\)
So we need to set \(\displaystyle{y}{\left({0}\right)}={0}\). But \(\displaystyle{y}'{\left({x}\right)}\) is a little bit harder t ocompute, I leave that to you.
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