FP1 - First order differential equations!

Hi,

I'm stuck on the following questions and would really appreciate some help:

1) Find the general solution of the differential equation:

dy/dx + 2ytanx = sinx

I got the integrating factor to be sec²x

Multiplying through by sec²x

sec²x dy/dx + 2ytanxsec²x = sinxsec²x

ysec²x = ∫ sinxsec²x dx

I've tried ∫ sinxsec²x so many times but can't end up with the answer which is secx +c. How do I get it?

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2) Find the general solution of the differential equation:

dy/dx + y/2x = -x^(1/2)

I rewrote this as dy/dx + (1/2x)y = -x^(1/2) to get it in the right form

Integrating factor = e^(∫ 1/2x dx) = e^(1/2ln|2x|) = |2x|^(1/2)

However, this doesn't seem right as the final answer comes out as
yx^(1/2)=C -1/2x² so the integrating factor must have been x^(1/2). I've checked my work a few times but can't see where I've gone wrong!

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3) How do I get from -1/8cos4x - 1/4cos2x + C to 3/2cos²x - cos^(4)x + C

It's for dy/dx + ycotx = cos3x

After the working out I get ysinx = ∫ sinxcos3x = 1/2 ∫ sin4x + sin2x although I'm not sure how to get the answer after what I got.

Thank you very much! :smile:

Reply 1

Speleo

INT sinx sec^2x = INT sinx/cos^2x = 1/cosx + C = secx + C
Use a substitution u = cosx if you can't see it.

e^INT 0.5/xdx = e^0.5INT1/xdx = e^0.5lnx = (e^lnx)^0.5 = x^0.5

e^(1/2ln(2x)) = e^(1/2lnx + 1/2ln2) = root2 * x^0.5 which is as above with a different constant of integration in the first part.

Reply 2

sweet_gurl

Thank you very much!

Can anyone please help me with question 3?

Reply 3

Speleo

sinx cos3x = sinx (cos2xcosx - sin2xsinx)
= (2cos^2x-1)(sinxcosx) - 2sin^3xcosx
= 2cos^3xsinx - sinxcosx - 2sin^3xcosx
Integrate to:
-1/2cos^4x + 1/2cos^2x - 1/2sin^4x + C
= -1/2cos^4x + 1/2cos^2x - 1/2(1-cos^2x)(1-cos^2x) + C
= -1/2cos^4x + 1/2cos^2x - 1/2 + cos^2x - 1/2cos^4x + C
= -cos^4x + 3/2cos^2x - 1/2 + C
Just change the constant of integration to get the required result.

Reply 4

sweet_gurl

Thanks again for the help.

I was trying to work out whether I can from -1/8cos4x - 1/4cos2x + C to 3/2cos²x - cos^(4)x + C after simplifying, although it doesn't seem to work.
I thought it would be easier to put ∫ sinxcos3x = 1/2 ∫ sin4x + sin2x which gives -1/8cos4x - 1/4cos2x + C.

However, I can't quite get 3/2cos²x - cos^(4)x + C from there. Where am I going wrong?