Complete Question
The complete question is shown on the first uploaded image
Answer:
The solution is [tex]y^2 = \frac{1}{ 3 - 2 (1 + x^2 ) ^{\frac{1}{2} }}[/tex]
Step-by-step explanation:
From the question we are told that
The function is [tex]\left \{ {{y' = xy^3 (1 + x^2)^{-\frac{1}{2} }} \atop {y(0) = 1}} \right.[/tex]
Generally the above equation can be represented as
[tex]\frac{dy}{dx} = xy^3 (1+ x^2)^{- \frac{1}{2} }[/tex]
[tex]=> \frac{dy}{y^3} = \frac{x}{ (1 + x^2)^{\frac{1}{2} }} dx[/tex]
[tex]\int\limits { \frac{dy }{y^3} } \, = \int\limits {\frac{x}{(1 + x^2) ^{\frac{1}{2} }} } \, dx[/tex]
=> [tex]- \frac{1}{2y^2} = (1 + x^2)^{\frac{1}{2} }+ C[/tex]
From the question we are told that at y(0) = 1
[tex]- \frac{ 1}{2 * (1) } = (1 + (0)^2)^{\frac{1}{2} } + C[/tex]
[tex]C = - \frac{3}{2}[/tex]
So
[tex]- \frac{1}{2y^2} = (1 + x^2)^{- \frac{1}{2}} - \frac{3}{2}[/tex]
[tex]\frac{1}{y^2} = 3-2(1 + x^2)^{\frac{1}{2} }[/tex]
[tex]y^2 = \frac{1}{ 3 - 2 (1 + x^2 ) ^{\frac{1}{2} }}[/tex]
