This is review for me, but something is going awry

[Jamin2112]

Homework Statement

Find the solution of the following initial value problems:

(a) y'' - 2y' + 5y = 0, y(π/2)=0, y'(π/2)=0

Homework Equations

Just know to "guess" y=Ce^rt, how to bring in Euler's formula, and that sin(x) is odd while cos(x) is even

The Attempt at a Solution

Guess: y= e^rt

-----> r²e^rt - 2re^rt + 5e^rt = 0
-----> r² - 2r + 5 = 0
-----> (r-1)² = -5 + 1
-----> r - 1 = +/- √-4
-----> r = 1 +/- 2i
-----> y = e^t( C₁(cos2t+isin2t) + C₂(cos(-2t)+isin(-2t)) )
-----> y = e^t( C₁cos2t + C₁sin2t - C₂cos2t + C₂isin2t)
-----> y = e^t(K₁cos2t + K₂sin2t), where K₁, K₂ are new constants to represent (C₁ - C₂), (iC₁ + iC₂)But then come the initial conditions!

0 = e^π/2K₁(-1)
2 = K₁e^π/2(-1)+ 2K₂e^π/2(-1)

It just seems like a ghey problem if K₁ disappears

 

[Dick]

Looks good so far. Did you mean y'(pi/2)=2? What's so hard about the initial conditions? What are sin(pi) and cos(pi)?

 

[Jamin2112]

Oh, and another one.

y'' + 2y' + y = 3e^-t

I'll find the homogenous solution first. The characteristic polynomial is r² + 2r + 1 = 0 -----> r = -1, a repeated root ----> The homogenous solution is (C₁ + C₂t)e^-t.

Now I'll guess a second general solution, C₃e^-t.

------> C₃e^t - 2C3e^-t + C₃e^-t = ₃e^-t
------> 0 = 0.

Fork! So is the homogeneous solution all? Just for the heck of it, I tried y(t)=(At² + Bt + C)e^-t, but to no avail; there doesn't seem to be a polynomial that can be multiplied by e^-t without my getting 0=0.

Where should I go from here?

 

[Dick]

There is an inhomogeneous solution of the form (At^2+Bt+C)*exp(-t). You just missed it.

 

[Jamin2112]

Looks good so far. Did you mean y'(pi/2)=2? What's so hard about the initial conditions? What are sin(pi) and cos(pi)?

Well, yeah.

y(t) = C₁e^tcos2t + C₂e^tsin2t -----> y'(t) = et(-C₁2sin2t + C₁cos2t + 2C₂cost + C₂sin2t).

So,
0 = C₁e^π/2cos(π) + C₂e^π/2sin(π)
2 = -2C₁e^π/2sin(π) + C₁e^π/2cos(π) + 2C₂e^π/2cos(π) + C₂e^π/2sin(π)

----->

2 = -C₂e^π/2


----> y(t) = -e^-π/2e^tcos2t

 

[Jamin2112]

There is an inhomogeneous solution of the form (At^2+Bt+C)*exp(-t). You just missed it.

Allow me to demonstrate.

y(t) = At²e^-t + Bte^-t + Ce^-t
----> y'(t)= -At²e^-t + 2Ate^-t -Bte^-t + Be^-t - Ce^-t

----> y''(t)= At²e^-t - 2Ate^-t - 2Ate^-t + 2Ae^-t + Bte^-t -Be^-t - Be^-t + Ce^-t

-----> At²e^-t - 2Ate^-t - 2Ate^-t + 2Ae^-t + Bte^-t -Be^-t - Be^-t + Ce^-t + 2*[ -At²e^-t + 2Ate^-t -Bte^-t + Be^-t - Ce^-t ] + At²e^-t + Bte^-t + Ce^-t = 3e^-t

Everything on the left side cancels out!

 

[Dick]

Well, yeah.

y(t) = C₁e^tcos2t + C₂e^tsin2t -----> y'(t) = et(-C₁2sin2t + C₁cos2t + 2C₂cost + C₂sin2t).

So,
0 = C₁e^π/2cos(π) + C₂e^π/2sin(π)
2 = -2C₁eπ/2sin(π) + C₁e^π/2cos(π) + 2C₂e^π/2cos(π) + C₂e^π/2sin(π)

----->

2 = -C₂e^π/2


----> y(t) = -e^-π/2e^tcos2t

Yes, what's wrong or 'ghey' about that?

 

[Jamin2112]

Yes, what's wrong or 'ghey' about that?

I'm used to easy system of equations in these problems.

Something like

0 = C₁ + C₂
2 = 3C₁ - C₂,

not something with e^π/2

 

[Dick]

Allow me to demonstrate.

y(t) = At²e^-t + Bte^-t + Ce^-t
----> y'(t)= -At²e^-t + 2Ate^-t -Bte^-t + Be^-t - Ce^-t

----> y''(t)= At²e^-t - 2Ate^-t - 2Ate^-t + 2Ae^-t + Bte^-t -Be^-t - Be^-t + Ce^-t

-----> At²e^-t - 2Ate^-t - 2Ate^-t + 2Ae^-t + Bte^-t -Be^-t - Be^-t + Ce^-t + 2*[ -At²e^-t + 2Ate^-t -Bte^-t + Be^-t - Ce^-t ] + At²e^-t + Bte^-t + Ce^-t = 3e^-t

Everything on the left side cancels out!

Look, you can forget about the B and C parts. You should KNOW they are going to cancel. They are part of the homogenous solution. Are you really going to make me check that? I will ask you, what canceled the 2*A*exp(-t)?

 

[Dick]

I'm used to easy system of equations in these problems.

Something like

0 = C₁ + C₂
2 = 3C₁ - C₂,

not something with e^π/2

exp(pi/2) is just another constant. Doesn't make it any harder than a constant like '3'.

 

[Jamin2112]

Look, you can forget about the B and C parts. You should KNOW they are going to cancel. They are part of the homogenous solution. Are you really going to make me check that? I will ask you, what canceled the 2*A*exp(-t)?

Let us see.

y(t) = At²e^-t,
y'(t) = -At²e^-t + 2Ate^-t,
y''(t)=At²e^-t - 2Ate^-t - 2Ate^-t + 2Ae^-t

------------>

At²e^-t - 2Ate^-t - 2Ate^-t + 2Ae^-t -2At²e^-t + 4Ate^-t + At²e^-t = 3e^-t


----------> A = 3/2

I think I finally got it. What a ghey problem.

WAIT! Now I have the homogeneous solution, in which the coefficients are undetermined, PLUS one another e^-t with a determined coefficient. How does that work? What was the point if (3/2 + C₁) could just be merged into a new constant?

 

[Dick]

Let us see.

y(t) = At²e^-t,
y'(t) = -At²e^-t + 2Ate^-t,
y''(t)=At²e^-t - 2Ate^-t - 2Ate^-t + 2Ae^-t

------------>

At²e^-t - 2Ate^-t - 2Ate^-t + 2Ae^-t -2At²e^-t + 4Ate^-t + At²e^-t = 3e^-t


----------> A = 3/2

I think I finally got it. What a ghey problem.

WAIT! Now I have the homogeneous solution, in which the coefficients are undetermined, PLUS one another e^-t with a determined coefficient. How does that work? What was the point if (3/2 + C₁) could just be merged into a new constant?

I would stop saying 'ghey'. I don't think it's universally appreciated. You can't absorb the inhomogeneous solution into the homogeneous solution by adjusting a constant. You are just being sloppy again.