 # 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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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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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.