Solve sine^x Variation of Parameters: y"+3y'+2y

[Timberhead]

Homework Statement

Solve by variation of parameters:
y" + 3y' + 2y = sine^x

Homework Equations

Finding the complimentary yields:
y_c = c₁e^-x + c₂e^-2x

The Attempt at a Solution

I set up the Wronskians and got:
μ₁ = ∫e^-2xsin(e^x)dx
μ₂ = -∫e^-xsin(e^x)dx

The problem is that I have no idea how to integrate sin(e^x).

I tried subbing u = e^-x du = -e^-x for μ₁
=> ∫-u du sin(u^-1)
Integration by parts either attempts to integrate sin(u^-1) or endlessly integrates u without repeat.

That failed, so I tried just integrating by parts of μ₁; it took 2 repetitions to get:
μ₁ = -½e^-2xsin(e^x) - ½e^-xcos(e^x) + ∫½sin(e^x)dx
I thought it might work if I get -e^x∫½sin(e^x)] in μ₂
μ₂ = e^-xsin(e^x) - ∫cos(e^x)dx

Going further into the integration by parts just adds more complications, such as adding "x" as a term as well as going into higher powers of e^x.

I can't express the integral as a series; that's next chapter and not covered on the mid-term in a few days (I'm currently hoping the mid-term doesn't have this problem).

 

[Simon Bridge]

Did you try substitution: $u=e^x$.

 

[Timberhead]

I did initially, but found it problematic since it ends up becoming ½∫^sinu/_u. If you keep on doing it it winds up being a giant beast of an equation which doesn't appear to match μ₂ at all.

I thought part of it might equal μ₂, but that didn't happen either. The two just keep mismatching cosine and sine functions endlessly.

I just realized I did the Wronskians wrong; the bottom isn't 1... And it ends up a LOT easier, but I still can't quite seem to get it right.
With the correct Wronskian I instead got ∫^sin(u)/_u for μ₁ and a nice and short -½cos(e^x) for μ₂. This does cause an elimination between the two equations, but I'm still getting a vast number of functions for μ₁ that doesn't seem to have an end.

I'm going to check back on this in the morning.

 

[Ray Vickson]

Homework Statement

Solve by variation of parameters:
y" + 3y' + 2y = sine^x

Homework Equations

Finding the complimentary yields:
y_c = c₁e^-x + c₂e^-2x

The Attempt at a Solution

I set up the Wronskians and got:
μ₁ = ∫e^-2xsin(e^x)dx
μ₂ = -∫e^-xsin(e^x)dx

The problem is that I have no idea how to integrate sin(e^x).

I tried subbing u = e^-x du = -e^-x for μ₁
=> ∫-u du sin(u^-1)
Integration by parts either attempts to integrate sin(u^-1) or endlessly integrates u without repeat.

That failed, so I tried just integrating by parts of μ₁; it took 2 repetitions to get:
μ₁ = -½e^-2xsin(e^x) - ½e^-xcos(e^x) + ∫½sin(e^x)dx
I thought it might work if I get -e^x∫½sin(e^x)] in μ₂
μ₂ = e^-xsin(e^x) - ∫cos(e^x)dx

Going further into the integration by parts just adds more complications, such as adding "x" as a term as well as going into higher powers of e^x.

I can't express the integral as a series; that's next chapter and not covered on the mid-term in a few days (I'm currently hoping the mid-term doesn't have this problem).

The integral $\int \sin(e^x) \, dx = \text{Si}(e^x) + C$, where $\text{Si}$ is a "non-elementary" function.

 

[Timberhead]

Resolved: I kept working with substituting e^x = u on both μ's and it actually works insanely well.

 

[Simon Bridge]

Well done - getting the correct Wronskian is helpful too of course ;)
(If the Wronskian was incorrect, does that make it a Wrightskian?)

Aside:
The "non-elementary function" Si(x) in post #4 is called the Sine Integral and you can look it up.
For that matter, sin(x)/x is called "sinc(x)" ... and you can look that up too.