# Non-homogeneous Equation; killing me!

#### Eng

Hey guys,

This is my first post here! Because I really stuck with this Non-homogeneous Eq.

y''-y'-2y=3e^(-t)

When I tried to solve it to find A, I always end up with:
2Ae^(-t)-2Ae^(-t) which gives ZERO=3e^(-t)

So, I couldn't solve for A since there is no A! Did I miss something? Plz help!

Thanx,

#### topsquark

Math Team
Hey guys,

This is my first post here! Because I really stuck with this Non-homogeneous Eq.

y''-y'-2y=3e^(-t)

When I tried to solve it to find A, I always end up with:
2Ae^(-t)-2Ae^(-t) which gives ZERO=3e^(-t)

So, I couldn't solve for A since there is no A! Did I miss something? Plz help!

Thanx,
This is happening because $$\displaystyle e^{-t}$$ is a solution to the homogeneous equation. Try $$\displaystyle t e^{-t}$$ as your trial particular solution.

-Dan

3 people

#### Country Boy

Math Team
In addition to the "method of undetermined coefficients", which only works when the "right-hand side" is one of the types of function we expect as solution to a homogeneous equation with constant coefficients (polynomials or exponentials or sine and cosine and combinations of those) and is more complicated if the "right-hand side" is a solution to the homogeneous equation, we can use "variation of parameters".

Having determined that the general solution to the associated homogeneous equation is $$\displaystyle y_h(t)= Ce^{-t}+ De^{2t}$$, we look for a solution to the entire equation of the form $$\displaystyle y_p(t)= u(t)e^{-t}+ v(t)e^{2t}$$ where u and v are some functions of t. It's easy to see that, whatever the true solution is, we can always find such functions - in fact there are many such solutions. Differentiating, $$\displaystyle y'_p(t)= u'(t)e^{-t}- u(t)e^{-t}+ v'(t)e^{2t}+ 2v(t)e^{2t}$$. Because there are many such solutions, we can "narrow the search", as well as simplify the equation, by requiring that $$\displaystyle u'(t)e^{-t}+ v'(t)e^{2t}= 0$$. Then we have $$\displaystyle y'_p(t)= -u(t)e^{-t}+ 2v(t)e^{2t}$$. Differentiating again, $$\displaystyle y''_p(t)= -u'(t)e^{-t}+ u(t)e^{-t}+ 2v'(t)e^{2t}+ 5v(t)e^{2t}$$. Putting those into $$\displaystyle y''- y'- 2y= 3e^{-t}$$, everything that does NOT involve u' or v' cancels (because $$\displaystyle e^{-t}$$ and $$\displaystyle e^{2t}$$ satisfy the associated homogeneous equation) so that we have $$\displaystyle -u'e^{-t}+ 2v'e^{2t}= 3e^{-t}$$.

So we have the two equations, $$\displaystyle u'e^{-t}+ v'e^{2t}= 0$$ and $$\displaystyle -u'e^{-t}+ 2v'e^{2t}= 3e^{-t}$$ that we can solve for u' and v'. Simply adding those two equations eliminates u': $$\displaystyle 3v'e^{2t}= 3e^{-t}$$ so $$\displaystyle v'= e^{-3t}$$ and then $$\displaystyle v(t)= -\frac{1}{3}e^{-3t}$$ (Since I am only seeking one solution, I can ignore the constant of integration). Since $$\displaystyle v'(t)= e^{-3t}$$, $$\displaystyle u'e^{-t}+ v'e^{2t}= u'e^{-t}+ e^{-t}= 0$$. So $$\displaystyle u'e^{-t}= -e^{-t}$$ and $$\displaystyle u'(t)= -1$$ so that $$\displaystyle u(t)= t$$ (That's where that "t" multiplying $$\displaystyle e^{-t}$$ in topsquark's post came from!). So $$\displaystyle y_p(t)= u(t)e^{-t}+ v(t)e^{2t}= te^{-t}-\frac{1}{3} e^{-3t}e^{2t}= te^{-t}- \frac{1}{3}e^{-t}$$. That last $$\displaystyle -\frac{1}{3}e^{-t}$$ can be absorbed into the "$$\displaystyle Ce^{-t}$$" part of the homogeneous solution giving $$\displaystyle y(t)= Ce^{-t}+ De^{2t}+ te^{-t}$$ as the general solution to the entire equation.

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2 people

#### v8archie

Math Team
It's easy to see that, whatever the true solution is, we can always find such functions- in fact there are many such solutions.
It's not easy to see this at all in my opinion. Please explain why you think it is.

It is also far from clear that your justification for requiring $u' \mathrm e^{-t} + v' \mathrm e^{2t}=0$ is valid. In particular, the claim that there even exists such a solution is not clear (until after we have produced it).

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1 person

#### v8archie

Math Team
Since I am only seeking one solution, I can ignore the constant of integration
The real reason you can ignore the constant of integration is that it represents a part of the complementary solution that you are about to add back in - so in reality you do not ignore it at all. Instead the $y_p$ is actually $y$, the general solution to the equation.

#### idontknow

search in $$\displaystyle y=[p_{r}(x)e^{ax}+p_{r}(x)e^{bx}]x^{r}$$
$$\displaystyle r$$ -- polynom degree
$$\displaystyle p_{r}(x)=e^{lnp_{r}(x)}$$ , $$\displaystyle y_1=e^{lnp_{r}(x)+c}x^{r}$$
$$\displaystyle y_2=e^{lnp_{r}(x)+c}$$ , $$\displaystyle y=(y_1+y_2+...+y_n+....)$$
the sum would give the same answer : $$\displaystyle ce^{-x}+de^{2x}+xe^{-x}$$

1 person

#### v8archie

Math Team
It's not easy to see this at all in my opinion. Please explain why you think it is.
I've come up with a justification as to why such functions exist, and why there are many of them. But it is far from clear why you believe this guarantees that one pair should have $u'\mathrm e^{-t} + v'\mathrm e^{2t}=0$.

Personally, I strongly dislike "it is easy to see" and "it is obvious", because frequently it isn't. And even if it is, the reader wonders what trick they might have missed. Also, if it's really that easy to see an explanation of one or two sentences would be easy to write and would avoid all confusion.

Usually I read it as "I'm sure of this, but I'm unable to show it".

#### skipjack

Forum Staff
$$\displaystyle y'' - y' - 2y = 3e^{-t}$$
$$\displaystyle e^{-2t}y'' - e^{-2t}y' - 2e^{-2t}y = 3e^{-3t}$$
$$\displaystyle e^{-2t}y' + e^{-2t}y = -e^{-3t} + 3\text{b}$$, where $\text{b}$ is a constant.
$$\displaystyle e^ty' + e^ty = -1 + 3\text{b}e^{3t}$$
$$\displaystyle e^ty = -t + \text{b}e^{3t} + \text{c}$$, where $\text{c}$ is a constant.
$$\displaystyle y = -te^{-t} + \text{b}e^{2t} + \text{c}e^{-t}$$

2 people

#### idontknow

btw , your title contains "killing me"
you can get killed only once, not continuously
Kill is an avaliable process only in Planck-Time in physics.
planck-time says ... $$\displaystyle t_p=\sqrt{\frac{\alpha G}{c^5}}\sim 5.39106 \times 10^{-44} s$$

#### v8archie

Math Team
Grammatically, use of the present continuous indicates that the action is in progress at the time of speaking or writing. That only requires that an action has a duration. In this case, the verb to kill meaning to extinguish the life of a living thing.

Biologically, the extinguishing of life happens over a period of time, not instantly. Indeed, there is little consensus among either medical professionals or scientists over when a living being can be pronounced dead. And consensus is even less common once lawyers get involved.

1 person