Question

In this next example, why do we add "A" to the RHS and not the LHS. I'm trying to understand nonhomogenous DE...sigh :-(
\[y''+4y=e^{3x}\]
...
\[y=Ae^{3x}\]
I feel like I'm just blindly following a pattern, but I wouldn't be able to recreate this on my own unless I memorize this pattern.

Answer 1

you're not adding it.
it's an arbitrary constant.

Answer 2

I guess I'm having a hard time expressing myself. I mean why is the A and such on the RHS and not the LHS?...What would I think of when I think of a nonhomogenous DE...what's my goal? Why am I deriving the RHS 2 times and then solving for A?

Answer 3

You're trying to find what function, y, makes the differential equation true.

Answer 4

But it seems like something's missing here . . .

Answer 5

would that mean that
\[4y=Ae^{3x}\]
\[y'=3Ae^{3x}=0\]
\[y''=9Ae^{3x}=1\]
?

Answer 6

Yes, something like that.
If y=Ae^{3x}, then y' = 3Ae^{3x}, and y'' = 9Ae^{3x}
Putting that into the original equation,
9Ae^{3x} +4(3Ae^{3x}) = e^{3x}

Answer 7

Yep. Do we then solve for A?

Answer 8

Yes.

Answer 9

this will allow us to get the equation y=.... which was our goal?

Answer 10

Yes, that will give what is called a particular solution. You'll need initial values / boundary conditions to get an exact solution for those conditions.

Answer 11

Do you know how to use the characteristic equation to find the complementary solution?

Answer 12

I guess it's coming together now.
but
\[y=e^{\alpha x}(c_1cos\beta x+c_2 sin\beta x)\]
because the auxiliary equation is
r==/- 2i

Answer 13

I meant +/-2i

Answer 14

auxiliary is \[r^2+4=0\]

Answer 15

initial values / boundary conditions....what r_1 and r_2 be considered?

Answer 16

Right. From the LHS you can treat it as a homogeneous equation to find the characteristic/auxilery, then combine that with the particular solution to get the general solution. Only then do you apply the initial conditions to determine your constants.

Answer 17

I'm sorry but what are my initial conditions?

Answer 18

did I ask a really bad question? :S

Answer 19

I don't know what they are. Do you have any? If you don't have any, then you leave it in general form and you're done (you'll still have undetermined coefficients, but if you don't know initial conditions, you just have to leave it that way.

Answer 20

Yeah I don't think there were any initial conditions. Ok I think I kinda get nonhomogenous DE's a little better now. I'll try a problem and we'll see how I do. I'll post it in a second.
Thanks soo much for all of your help and patience :-)

Answer 21

I can show you an example of an initial value problem if you like.

Answer 22

sure

Answer 23

Ok, I'll do a simple one so it'll be easy to follow.

Given: \[y''-4y'-12y=3e^{5t}, \space y(0)=\frac{18}{7}, \space y'(0)=-\frac{1}{7}\]

Answer 24

The general solution is:
\[y(t)=y_c(t)+Y_p(t)\]
Where y_c is the characteristic equation, and Y_p is the particular.

Answer 25

Sorry *complementary, not "characteristic"

Answer 26

The characteristic equation is found from the LHS:
\[r^2-4r-12=0\]
The solutions for r are {-2, 6}

Answer 27

This makes the complementary equation
\[y_c(t)=c_1e^{-2t}+c_2e^{6t}\]

Answer 28

(This should all look familiar, I hope)

Answer 29

yep

Answer 30

Ok.
The particular solution is a guess that y is something like e^x in order to be differentiated and still end up with something like e^x on the RHS, so we guess:
\[Y_p(t)=Ae^{5t}\]

Answer 31

Differentiate that once, then again, and plug y, y', and y'' into the original equation to solve for A. It looks like it'll come out to A=-3/7, but you should double-check that for me.

Answer 32

what happened to the 2 in front of the e?

Answer 33

i mean the 3

Answer 34

That was from the RHS, this is all still solving for the LHS. The 3 is still there when you substitute and solve for A:
\[(Ae^{5t})''-4(Ae^{5t})'-12(Ae^{5t})=3e^{5t}\]

Answer 35

oh ok...so I guess for the sake of solving for A we leave it out?

Answer 36

oh never mind. I see what you're doing

Answer 37

Cool. Can you solve that for A and verify that it is -3/7?

Answer 38

One more thing. So I it a general rule to say \[Y_P(t)=Ae^5t\]
and we're using e^5t because that's in the original equation?
ok now back to solving for A....

Answer 39

Time to put it all together.
The complementary solution is
\[y_c=c_1e^{-2t}+c_2e^{6t}\]
and the particular solution is
\[Y_p=-\frac{3}{7}e^{5t}\]
Altogether, the general solution is
\[y(t)=c_1e^{-2t}+c_2e^{6t}-\frac{3}{7}e^{5t}\]

Answer 40

and yes I got \[-\frac37\] for A

Answer 41

It's a general rule to guess that y looks like Ae^{bt} because that is what was on the RHS.
We know that you can differentiate Ae^{bt} as many times as you want and it'll still look pretty much like Ae^{bt}.

Answer 42

got it. It's all coming together now

Answer 43

If it was a different function, say, sin(x), then you'd want to guess something like sin(x) because the second derivative of sin(x) is -sin(x), so it only changed by a constant.

Answer 44

Alright, so, now that we have the general solution, the only thing left is to apply the initial conditions to find c_1 and c_2.

Answer 45

The initial conditions are in terms of y and y', so you'll have to differentiate our general y(t) to get that.

Answer 46

ok
\[\frac{18}7=c_1+c_2-\frac37\]
and
\[y'=-2c_1e^{-2t}+6c_2e^{6t}-\frac{15}7e^{5t}\]

Answer 47

still working on it :)

Answer 48

\[y'(0)=-\frac17=-2c_1+6c_2-\frac{15}{7}\]

Answer 49

so now we have two equations and 2 unknowns...ok lets see

Answer 50

\[c_2=41/14\]

Answer 51

and c_1=1/7

Answer 52

Hmm, that doesn't look right; let me double-check..

Answer 53

Your equations look right, but they simplify to
\[c_1+c_2=3\]
\[-2c_1+6c_2=2\]
Solving that system, c1=2 and c2=1

Answer 54

what? How did you do that?

Answer 55

c_1 + c_2 - 3/7 = 18/7 (from y_0)
-2c_1 + 6c_2 - 15/7 = -1/7 (from y'_0)

Those are the equations you got, right?
Try solving that system again.

Answer 56

I get -1 for c_2

Answer 57

oh sorry I fault

Answer 58

c_2 should be 2

Answer 59

well I'll look at it again I guess =P
gtg for now

Answer 60

Ok, so go through it all again and verify that the solution is
\[y(t)=2e^{-2t}+e^{6t}-\frac{3}{7}e^{5t}\]

Try taking the first and second derivatives of that and putting them back in the original equation to see if it's true and valid.