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Homework Help: Differential equation to homogenous linear equation.

  1. Jan 15, 2014 #1
    can this equation y, = ycot(x) + sin(x) be reduced to a homogenous linear format? If yes, how?
    I tried the usual y=xv and the x=X+h, y=Y+k but doesn't seem to be working. Any ideas?


    just realized its in the form of dx/dy+Px=Q so I solved it by multiplying on B.S. by e∫Pdx and the solution I got was y=cosec(x) - cos(x) + cot (x)/2, but this is not given in the options. Did I go wrong somewhere?
    Last edited: Jan 15, 2014
  2. jcsd
  3. Jan 15, 2014 #2
    Yes, you went wrong somewhere. We cannot say where since we don't know what you did in full details.
  4. Jan 15, 2014 #3
    Well, to reduce the equation to a homogeneous linear equation which is in the form of dy/dx + Py = Q we multiply it on both sides with e∫Pdx (at least to my understanding). In this case P = -cotx, hence the IF becomes e∫cot(x)dx which becomes e-ln|sin(x)| which then becomes just 1/sin(x) so we have y/sinx=x, that means y=xsinx. Is this right? and moreover I need to find the general solution with C as the integration constant and have no idea how to do that. Can you now tell me please where I went haywire?

    BTW JJacquelin, thanks coz after reading your comment I rechecked and realized that I made a huge mistake with the sign convention and redid the whole problem but I think I am almost there now. Just need the general solution with C as the integration constant.

  5. Jan 15, 2014 #4
    Wait, I integrated 1 which resulted in that x before sinx. That means any constant present there would be the integration constant and the general solution is to replace that x with C. So the general solution would be y = Csinx. Am I right?
  6. Jan 15, 2014 #5
    Your message #3 is OK : y=x*sin(x) is a particular solution.
    But, sorry to say, the message #4 is full of confusion. If you bring back y=Csin(x) into the equation y'=ycot(x)+sin(x) what do you obtain ? Is it correct ?
  7. Jan 17, 2014 #6
    If I put y=Csin(x) in y,=ycot(x)+sin(x) I get y,=Ccos(x)+sin(x)
    The question specifies that there is a general solution with C as the integration constant. Trouble is my understanding of integration constants says that the general solution should be (in this case) y=(x+C)*sin(x) because I integrated 1 at RHS which gives x+C and the denominator at the LHS was y/sin(x) taking it to the RHS we should have y=(x+C)*sin(x) (I don't know if this substitutes as a general solution) but the choices I have are
    a) y=cos(Cx)
    b) y=sin(Cx)
    c) y=Ccos(x)
    d) y=Csin(x)
    None of these have a "+C" as the integration constant, all are multiplied with C, how does that happen?
  8. Jan 18, 2014 #7


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    Last edited: Jan 18, 2014
  9. Jan 18, 2014 #8
    No, you didn't put it completely. If you put y=Csin(x) in y,=ycot(x)+sin(x) you get Ccos(x)=Ccos(x)+sin(x) because y'=Ccos(x). You see that the result is false because Ccos(x) is not equal to Ccos(x)+sin(x). This shows that y=Csin(x) is not the general solution of the equation y,=ycot(x)+sin(x)
    Now, put y=(x+C)*sin(x) in y,=ycot(x)+sin(x) and see if the equality is obtained or not. Then you will be able to conclude if y=(x+C)*sin(x) is the general solution or not.
  10. Jan 18, 2014 #9
    If I put y=(x+C)sinx then y,=xcosx+sinx+Ccosx and when I substitute in y,=ycotx+sinx we have xcosx+sinx+Ccosx on both sides. Hence equality is obtained. But it is still not among the given options? :cry::confused:
  11. Jan 18, 2014 #10


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    The given options are wrong.

  12. Jan 18, 2014 #11
    And I'll say even more: "All the given options are wrong" :devil:
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