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Math Help - First order ODEs

  1. #1
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    First order ODEs

    Greetings, I have a few differential equation questions and I began to take this course at the beginning of this semester so I might have been doing obvious mistakes.These questions are from Ross, Differential Equations 1st edition chapter 2.2.
    3.  2r(s^2+1)dr+(r^4+1)ds=0
    14. (\sqrt(x+y)+\sqrt(x-y))dx+(\sqrt(x-y)-\sqrt(x+y))dy=0
    23.(a)Prove that if Mdx+Ndy=0 is a homogeneous equation, then the change of variables x=uy transforms this equation into a separable equation in the variables u and x.
    (b)Use this result to solve (x^2-3y^2)dx+2xydy=0
    (c)Use this result to solve (y+\sqrt(x^2+y^2))dx-xdy=0

    In Q3 I found \arctan(r^2)+\arctan(s)=C and unable to reduce this expresion to r^2+s=c(1-r^2s) as given in the answers at the back of the book. In Q14 I got the following expression and unable to proceed further:
    \int(\frac{(\sqrt(1+v)-\sqrt(1-v))dv}{\sqrt(1+v)+\sqrt(1-v)+v\sqrt(1-v)-v\sqrt(1+v)})=\ln|x|+C where v=\frac{y}{x}.
    In Q23 I started with \frac{dy}{dx}=h(x/y) then letting x=uy leads \frac{dy}{dx}=\frac{1}{u}-\frac{x}{u^2}\frac{du}{dx}=h(u)
    It follows that:
    \int\frac{dx}{x}-\int\frac{du}{u^2(\frac{1}{u}-h(u))}=c
    Is the aforementioned approach right or wrong?

    Any help is appreciated,

    Thanks in advance.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: First order ODEs

    Q3:
    I also get \arctan(r^2)=-\arctan(s)+C \Leftrightarrow \arctan(r^2)+\arctan(s)=C
    Now, I would do the following:
    \tan[ \arctan(r^2)+\arctan(s)]=\tan(C)

    Expand the LHS by using \tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\cdot \tan(b)} and using the fact that \tan[\arctan(x)]=x
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  3. #3
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    Re: First order ODEs

    Thanks Siron, I managed to understand Q3. Do you have any idea about the other questions especially Q14?
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  4. #4
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    Re: First order ODEs

    For 14 you may try

    x+y=p

    x-y=q.
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  5. #5
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    Re: First order ODEs

    Thanks, I will try the substitution you mentioned. Any other suggestions?
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