Homework Help: Differential equation to homogenous linear equation.

1. Jan 15, 2014

JonNash

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?

Thanks

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. Jan 15, 2014

JJacquelin

Yes, you went wrong somewhere. We cannot say where since we don't know what you did in full details.

3. Jan 15, 2014

JonNash

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.

Thanks.

4. Jan 15, 2014

JonNash

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?

5. Jan 15, 2014

JJacquelin

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 ?

6. Jan 17, 2014

JonNash

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?

7. Jan 18, 2014

ehild

Last edited: Jan 18, 2014
8. Jan 18, 2014

JJacquelin

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.

9. Jan 18, 2014

JonNash

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?

10. Jan 18, 2014

ehild

The given options are wrong.

ehild

11. Jan 18, 2014

JJacquelin

And I'll say even more: "All the given options are wrong"