Question

Find the solution (in explicit form) for the following initial value problem:    y 0 = xy3 (1 + x 2 ) − 1 2 y(0) = 1.

Answer 1

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The solution is  $y^2 = \frac{1}{ 3 - 2 (1 + x^2 ) ^{\frac{1}{2} }}$    

Step-by-step explanation:

From the question we are told that

  The  function is  $\left \{ {{y' = xy^3 (1 + x^2)^{-\frac{1}{2} }} \atop {y(0) = 1}} \right.$

Generally the above  equation can be represented as

      $\frac{dy}{dx} = xy^3 (1+ x^2)^{- \frac{1}{2} }$

      $=> \frac{dy}{y^3} = \frac{x}{ (1 + x^2)^{\frac{1}{2} }} dx$

     $\int\limits { \frac{dy }{y^3} } \, = \int\limits {\frac{x}{(1 + x^2) ^{\frac{1}{2} }} } \, dx$

=>    $- \frac{1}{2y^2} = (1 + x^2)^{\frac{1}{2} }+ C$

From the question we are told that at y(0) =  1

     $- \frac{ 1}{2 * (1) } = (1 + (0)^2)^{\frac{1}{2} } + C$

     $C = - \frac{3}{2}$

So

       $- \frac{1}{2y^2} = (1 + x^2)^{- \frac{1}{2}} - \frac{3}{2}$      

     $\frac{1}{y^2} = 3-2(1 + x^2)^{\frac{1}{2} }$

     $y^2 = \frac{1}{ 3 - 2 (1 + x^2 ) ^{\frac{1}{2} }}$