Finding the general solution of the given D.E.

[shamieh]

Find the general solution of the given differential equation.

$y'' + 2y' + y = 2e^{-t}$

I understand how to do everything in the problem- but I do not understand how we are obtaining $y = At^2e^{-t}$? Is this just some formula I should memorize or how exactly am I supposed to guess or figure out what $y$ is? I solved the characteristic equation and got $c_1e^{-t} + c_2te^{-t}$ (repeated root eqn)..The book then implies that since we know the characteristic eqn is $c_1e^{-t} + c_2te^{-t}$ then we must assume $y = At^2e^{-t}$.. How in the world are they assuming that? What is the formula that would drive them to knowing that? I'm really confused.

 

[MarkFL]

Your characteristic roots would be complex, so the general solution would be different. Is it possible the given ODE is:

\(\displaystyle y''+2y'+y=2e^{-t}\)

If this is in fact the case, let's first look at multiplying through by \(\displaystyle e^{t}\):

\(\displaystyle e^{t}y''+2e^{t}y'+e^{t}y=2\)

Notice now we can write:

\(\displaystyle \frac{d^2}{dt^2}\left(e^{t}y\right)=2\)

Integrating with respect to $t$, we obtain:

\(\displaystyle \frac{d}{dt}\left(e^{t}y\right)=2t+c_1\)

Integrating again with respect to $t$, we obtain:

\(\displaystyle e^{t}y=t^2+c_1t+c_2\)

Hence, the general solution is:

\(\displaystyle y(t)=t^2e^{-t}+c_1te^{-t}+c_2e^{-t}=e^{-t}\left(t^2+c_1t+c_2\right)\)

 

[shamieh]

Your characteristic roots would be complex, so the general solution would be different. Is it possible the given ODE is:

\(\displaystyle y''+2y'+y=2e^{-t}\)

If this is in fact the case, let's first look at multiplying through by \(\displaystyle e^{t}\):

\(\displaystyle e^{t}y''+2e^{t}y'+e^{t}y=2\)

Notice now we can write:

\(\displaystyle \frac{d^2}{dt^2}\left(e^{t}y\right)=2\)

Integrating with respect to $t$, we obtain:

\(\displaystyle \frac{d}{dt}\left(e^{t}y\right)=2t+c_1\)

Integrating again with respect to $t$, we obtain:

\(\displaystyle e^{t}y=t^2+c_1t+c_2\)

Hence, the general solution is:

\(\displaystyle y(t)=t^2e^{-t}+c_1te^{-t}+c_2e^{-t}=e^{-t}\left(t^2+c_1t+c_2\right)\)

You are correct Mark, forgot to type 2 in unfortunately. Reading your remarks currently to make sure I understand.

 

[shamieh]

Wow, I see what you did. Your method definitely works.. I got the same answer, but for some reason the book wanted me to take derivatives after I found what $y$ was and then plug it back into my eqn and solve for $A$ which I got to be $1$. Long story short, multiply by $e^t$ (i.e. your method) is like 2 steps versus this god awful book which made me plug that mess back into the equation using tedious steps. Thank!

 

[MarkFL]

Wow, I see what you did. Your method definitely works.. I got the same answer, but for some reason the book wanted me to take derivatives after I found what $y$ was and then plug it back into my eqn and solve for $A$ which I got to be $1$. Long story short, multiply by $e^t$ (i.e. your method) is like 2 steps versus this god awful book which made me plug that mess back into the equation using tedious steps. Thank!

Yes, your book likely wanted you to use the method of undetermined coefficients, just as I was taught as a student. The method I used above I have since learned from another user on the forums. :D

What your book wanted you to do is observe you have the characteristic root $r=-1$ of multiplicity 2, hence the homogeneous solution is:

\(\displaystyle y_h(t)=c_1e^{-t}+c_2te^{-t}\)

Then, we need a particular solution such that no term in the particular solution is a solution to the corresponding homogenous equation, thus we must assume a particular solution of the form:

\(\displaystyle y_p(t)=At^2e^{-t}\)

Hence:

\(\displaystyle y_p'(t)=A\left(2t-t^2\right)e^{-t}\)

\(\displaystyle y_p''(t)=A\left(t^2-4t+2\right)e^{-t}\)

Now, plugging the particular solution into the original ODE, we obtain:

\(\displaystyle A\left(t^2-4t+2\right)e^{-t}+2A\left(2t-t^2\right)e^{-t}+At^2e^{-t}=2e^{-t}\)

Multiply through by $e^{t}$ and arrange as:

\(\displaystyle 2A=2\implies A=1\)

Thus

\(\displaystyle y_p(t)=t^2e^{-t}\)

And finally by the principle of superposition, we may write:

\(\displaystyle y(t)=y_h(t)+y_p(t)=c_1e^{-t}+c_2te^{-t}+t^2e^{-t}\)

This is equivalent to the solution obtained above. The method I used above is far less general than using undetermined coefficients, and just happened to work nicely for this problem. :D

 

[shamieh]

Hey Mark when we are picking a particular solution that doesn't relate to the homogeneous does it mean I can essentially choose anything? For Example you Let $y_p = At^2e^{-t}$ But could I let $y_p = At^3e^{-t}$ and still get a correct solution?

 

[MarkFL]

Hey Mark when we are picking a particular solution that doesn't relate to the homogeneous does it mean I can essentially choose anything? For Example you Let $y_p = At^2e^{-t}$ But could I let $y_p = At^3e^{-t}$ and still get a correct solution?

Try it and see what happens. :D

I was taught to assume a particular solution of the form (for this problem type)

\(\displaystyle y_p(t)=At^se^{-t}\)

where $s$ is the smallest non-negative integer such that $y_p$ has no terms that are a solution to the corresponding homogeneous equation. For this problem, we find $s=2$. To be perfectly honest, I don't recall what happens if we choose $s$ to be larger than necessary. But, explore it and see what you get...(Yes)