Use an appropriate change of variables to solve the differential equation.

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Use an appropriate change of variables to solve the differential equation.



$$tfracdydt-y=sqrtt^2+y^2$$



My friend and I are trying to figure out how to solve this equation. Our professor has given us several methods but we aren't sure which to use because none of the equations are similar to this one.



Any help would be appreciated especially if you could help us with step by step.



Thanks!



--



UPDATE:



$$tfracdydt-y=sqrtt^2+y^2$$



$$fracdydt-fracyt=sqrt1+fracy^2t$$ where u= y/t



$$fracdydt=sqrt1+u^2+u=f(u)$$



$$f(u)-u=sqrt1+u^2+u-u=sqrt1+u^2$$
...



We ended up with $ln|sqrt1+u^2|=ln|x|+c$






differential-equations derivatives change-of-variable







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  • As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
    – b00n heT
    1 hour ago















up vote
1
down vote

favorite












Use an appropriate change of variables to solve the differential equation.



$$tfracdydt-y=sqrtt^2+y^2$$



My friend and I are trying to figure out how to solve this equation. Our professor has given us several methods but we aren't sure which to use because none of the equations are similar to this one.



Any help would be appreciated especially if you could help us with step by step.



Thanks!



--



UPDATE:



$$tfracdydt-y=sqrtt^2+y^2$$



$$fracdydt-fracyt=sqrt1+fracy^2t$$ where u= y/t



$$fracdydt=sqrt1+u^2+u=f(u)$$



$$f(u)-u=sqrt1+u^2+u-u=sqrt1+u^2$$
...



We ended up with $ln|sqrt1+u^2|=ln|x|+c$






differential-equations derivatives change-of-variable







share|cite|improve this question























  • As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
    – b00n heT
    1 hour ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Use an appropriate change of variables to solve the differential equation.



$$tfracdydt-y=sqrtt^2+y^2$$



My friend and I are trying to figure out how to solve this equation. Our professor has given us several methods but we aren't sure which to use because none of the equations are similar to this one.



Any help would be appreciated especially if you could help us with step by step.



Thanks!



--



UPDATE:



$$tfracdydt-y=sqrtt^2+y^2$$



$$fracdydt-fracyt=sqrt1+fracy^2t$$ where u= y/t



$$fracdydt=sqrt1+u^2+u=f(u)$$



$$f(u)-u=sqrt1+u^2+u-u=sqrt1+u^2$$
...



We ended up with $ln|sqrt1+u^2|=ln|x|+c$






differential-equations derivatives change-of-variable







share|cite|improve this question















Use an appropriate change of variables to solve the differential equation.



$$tfracdydt-y=sqrtt^2+y^2$$



My friend and I are trying to figure out how to solve this equation. Our professor has given us several methods but we aren't sure which to use because none of the equations are similar to this one.



Any help would be appreciated especially if you could help us with step by step.



Thanks!



--



UPDATE:



$$tfracdydt-y=sqrtt^2+y^2$$



$$fracdydt-fracyt=sqrt1+fracy^2t$$ where u= y/t



$$fracdydt=sqrt1+u^2+u=f(u)$$



$$f(u)-u=sqrt1+u^2+u-u=sqrt1+u^2$$
...



We ended up with $ln|sqrt1+u^2|=ln|x|+c$






differential-equations derivatives change-of-variable




differential-equations derivatives change-of-variable






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share|cite|improve this question













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edited 54 mins ago

























asked 1 hour ago









Taljana D

180110




180110











  • As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
    – b00n heT
    1 hour ago

















  • As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
    – b00n heT
    1 hour ago
















As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
– b00n heT
1 hour ago





As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
– b00n heT
1 hour ago











4 Answers
4






active

oldest

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up vote
4
down vote













Divide by $t$ to obtain,
beginequation
fracdydt - fracyt = sqrt1+left ( fracyt right )^2,
endequation
then use a change of variables $u = fracyt$. The transformed equation should be integrable using standard methods.






share|cite|improve this answer








New contributor




Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Could you check what we've done above please!
    – Taljana D
    53 mins ago

















up vote
2
down vote













HINT



One easy one to try is $y=zt$ so $y'=z't+z$ and the ODE becomes $$z't = sqrt1+z^2$$ which is separable.






share|cite|improve this answer




















  • Can you check what we've done above please.
    – Taljana D
    52 mins ago

















up vote
2
down vote













Hint: Substitute $$y(t)=tv(t)$$ then you will get
$$tleft(tfracdv(t)dt+v(t)right)-tv(t)=sqrtt^2+t^2v(t)^2$$
Can you proceed?






share|cite|improve this answer




















  • Can you check what we've done above please!
    – Taljana D
    52 mins ago

















up vote
1
down vote













By using the substitution $displaystyle u=frac yt$ - and therefore $displaystyle fracdudt=fracty'-yt^2$ - the differential equation becomes



$$fracdudt=frac1tsqrt1+u^2$$



which can be solved by separation of variables. To be exact



$$beginalign
fracdudt&=frac1tsqrt1+u^2\
fracdusqrt1+u^2&=fracdtt\
intfracdusqrt1+u^2&=intfracdtt\
operatornamearsinh(u)&=log t +c\
u&=sinh(log t+c)
endalign$$



and therefore by resubstitution $y(t)=tcdotsinh(log t+c)$.






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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    4
    down vote













    Divide by $t$ to obtain,
    beginequation
    fracdydt - fracyt = sqrt1+left ( fracyt right )^2,
    endequation
    then use a change of variables $u = fracyt$. The transformed equation should be integrable using standard methods.






    share|cite|improve this answer








    New contributor




    Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.

















    • Could you check what we've done above please!
      – Taljana D
      53 mins ago














    up vote
    4
    down vote













    Divide by $t$ to obtain,
    beginequation
    fracdydt - fracyt = sqrt1+left ( fracyt right )^2,
    endequation
    then use a change of variables $u = fracyt$. The transformed equation should be integrable using standard methods.






    share|cite|improve this answer








    New contributor




    Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.

















    • Could you check what we've done above please!
      – Taljana D
      53 mins ago












    up vote
    4
    down vote










    up vote
    4
    down vote









    Divide by $t$ to obtain,
    beginequation
    fracdydt - fracyt = sqrt1+left ( fracyt right )^2,
    endequation
    then use a change of variables $u = fracyt$. The transformed equation should be integrable using standard methods.






    share|cite|improve this answer








    New contributor




    Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    Divide by $t$ to obtain,
    beginequation
    fracdydt - fracyt = sqrt1+left ( fracyt right )^2,
    endequation
    then use a change of variables $u = fracyt$. The transformed equation should be integrable using standard methods.







    share|cite|improve this answer








    New contributor




    Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    share|cite|improve this answer



    share|cite|improve this answer






    New contributor




    Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    answered 1 hour ago









    Andrew Krause

    1412




    1412




    New contributor




    Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





    New contributor





    Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.











    • Could you check what we've done above please!
      – Taljana D
      53 mins ago
















    • Could you check what we've done above please!
      – Taljana D
      53 mins ago















    Could you check what we've done above please!
    – Taljana D
    53 mins ago




    Could you check what we've done above please!
    – Taljana D
    53 mins ago










    up vote
    2
    down vote













    HINT



    One easy one to try is $y=zt$ so $y'=z't+z$ and the ODE becomes $$z't = sqrt1+z^2$$ which is separable.






    share|cite|improve this answer




















    • Can you check what we've done above please.
      – Taljana D
      52 mins ago














    up vote
    2
    down vote













    HINT



    One easy one to try is $y=zt$ so $y'=z't+z$ and the ODE becomes $$z't = sqrt1+z^2$$ which is separable.






    share|cite|improve this answer




















    • Can you check what we've done above please.
      – Taljana D
      52 mins ago












    up vote
    2
    down vote










    up vote
    2
    down vote









    HINT



    One easy one to try is $y=zt$ so $y'=z't+z$ and the ODE becomes $$z't = sqrt1+z^2$$ which is separable.






    share|cite|improve this answer












    HINT



    One easy one to try is $y=zt$ so $y'=z't+z$ and the ODE becomes $$z't = sqrt1+z^2$$ which is separable.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    gt6989b

    31.1k22349




    31.1k22349











    • Can you check what we've done above please.
      – Taljana D
      52 mins ago
















    • Can you check what we've done above please.
      – Taljana D
      52 mins ago















    Can you check what we've done above please.
    – Taljana D
    52 mins ago




    Can you check what we've done above please.
    – Taljana D
    52 mins ago










    up vote
    2
    down vote













    Hint: Substitute $$y(t)=tv(t)$$ then you will get
    $$tleft(tfracdv(t)dt+v(t)right)-tv(t)=sqrtt^2+t^2v(t)^2$$
    Can you proceed?






    share|cite|improve this answer




















    • Can you check what we've done above please!
      – Taljana D
      52 mins ago














    up vote
    2
    down vote













    Hint: Substitute $$y(t)=tv(t)$$ then you will get
    $$tleft(tfracdv(t)dt+v(t)right)-tv(t)=sqrtt^2+t^2v(t)^2$$
    Can you proceed?






    share|cite|improve this answer




















    • Can you check what we've done above please!
      – Taljana D
      52 mins ago












    up vote
    2
    down vote










    up vote
    2
    down vote









    Hint: Substitute $$y(t)=tv(t)$$ then you will get
    $$tleft(tfracdv(t)dt+v(t)right)-tv(t)=sqrtt^2+t^2v(t)^2$$
    Can you proceed?






    share|cite|improve this answer












    Hint: Substitute $$y(t)=tv(t)$$ then you will get
    $$tleft(tfracdv(t)dt+v(t)right)-tv(t)=sqrtt^2+t^2v(t)^2$$
    Can you proceed?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    Dr. Sonnhard Graubner

    69.6k32761




    69.6k32761











    • Can you check what we've done above please!
      – Taljana D
      52 mins ago
















    • Can you check what we've done above please!
      – Taljana D
      52 mins ago















    Can you check what we've done above please!
    – Taljana D
    52 mins ago




    Can you check what we've done above please!
    – Taljana D
    52 mins ago










    up vote
    1
    down vote













    By using the substitution $displaystyle u=frac yt$ - and therefore $displaystyle fracdudt=fracty'-yt^2$ - the differential equation becomes



    $$fracdudt=frac1tsqrt1+u^2$$



    which can be solved by separation of variables. To be exact



    $$beginalign
    fracdudt&=frac1tsqrt1+u^2\
    fracdusqrt1+u^2&=fracdtt\
    intfracdusqrt1+u^2&=intfracdtt\
    operatornamearsinh(u)&=log t +c\
    u&=sinh(log t+c)
    endalign$$



    and therefore by resubstitution $y(t)=tcdotsinh(log t+c)$.






    share|cite|improve this answer


























      up vote
      1
      down vote













      By using the substitution $displaystyle u=frac yt$ - and therefore $displaystyle fracdudt=fracty'-yt^2$ - the differential equation becomes



      $$fracdudt=frac1tsqrt1+u^2$$



      which can be solved by separation of variables. To be exact



      $$beginalign
      fracdudt&=frac1tsqrt1+u^2\
      fracdusqrt1+u^2&=fracdtt\
      intfracdusqrt1+u^2&=intfracdtt\
      operatornamearsinh(u)&=log t +c\
      u&=sinh(log t+c)
      endalign$$



      and therefore by resubstitution $y(t)=tcdotsinh(log t+c)$.






      share|cite|improve this answer
























        up vote
        1
        down vote










        up vote
        1
        down vote









        By using the substitution $displaystyle u=frac yt$ - and therefore $displaystyle fracdudt=fracty'-yt^2$ - the differential equation becomes



        $$fracdudt=frac1tsqrt1+u^2$$



        which can be solved by separation of variables. To be exact



        $$beginalign
        fracdudt&=frac1tsqrt1+u^2\
        fracdusqrt1+u^2&=fracdtt\
        intfracdusqrt1+u^2&=intfracdtt\
        operatornamearsinh(u)&=log t +c\
        u&=sinh(log t+c)
        endalign$$



        and therefore by resubstitution $y(t)=tcdotsinh(log t+c)$.






        share|cite|improve this answer














        By using the substitution $displaystyle u=frac yt$ - and therefore $displaystyle fracdudt=fracty'-yt^2$ - the differential equation becomes



        $$fracdudt=frac1tsqrt1+u^2$$



        which can be solved by separation of variables. To be exact



        $$beginalign
        fracdudt&=frac1tsqrt1+u^2\
        fracdusqrt1+u^2&=fracdtt\
        intfracdusqrt1+u^2&=intfracdtt\
        operatornamearsinh(u)&=log t +c\
        u&=sinh(log t+c)
        endalign$$



        and therefore by resubstitution $y(t)=tcdotsinh(log t+c)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 1 hour ago

























        answered 1 hour ago









        mrtaurho

        1,190320




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