I've been working on this question but just can't crack it. It is as follows.Using d y d x = a x 2 + y 2 + b y b x where y(1)=0and by letting v = y x transform the above differential equation to an equation in v and x. Find the implicit solution of the resulting separable equation. Then use the relationship y=xv to obtain y + x 2 + y 2 = x a b + 1 I've tried separate the upper terms into two fractions which makes the second term simply b y b x = v but I am still left with the nasty square root with the y on the inside and I can't shift it easily to solve as a first order DE. Also I was thinking about solving it as a linear first order ODE but we have a squared term. Sorry about formatting as well, first time posting a question.

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I&#039;ve been working on this question but just can&#039;t crack it. It is as follows.Using             d      y              d      x        =            a                        x          2                +                  y          2                    +      b      y              b      x       where y(1)=0and by letting   v  =      y    x   transform the above differential equation to an equation in v and x. Find the implicit solution of the resulting separable equation. Then use the relationship y=xv  to obtain   y  +            x      2        +          y      2        =      x                  a        b            +      1      I&#039;ve tried separate the upper terms into two fractions which makes the second term simply             b      y              b      x        =  v but I am still left with the nasty square root with the y on the inside and I can&#039;t shift it easily to solve as a first order DE. Also I was thinking about solving it as a linear first order ODE but we have a squared term. Sorry about formatting as well, first time posting a question.

assumintdz · Accepted Answer

Just as you did            d      y              d      x        =            a                        x          2                +                  y          2                    +      b      y              b      x      Let  y  =  x  v    ⟹        y    ′    =  v  +  x      v    ′  making the equation to be  x      v    ′    =      a    b        1    +          v      2      which is separable  ∫            d      x        x    =      b    a    ∫            d      v              1      +              v        2            making  log  ⁡  (  x  )  +  C  =      b    a        sinh          −      1        ⁡  (  v  )So  v  =  sinh  ⁡      (          a      b        log    ⁡    (    x    )    +    C    )      ⟹    y  =  x    sinh  ⁡      (          a      b        log    ⁡    (    x    )    +    C    )

I've been working on this question but just can't crack it. It is as follows.Using...

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