Respuesta :
Answer:
[tex]y = \dfrac{2}{3-x^2}[/tex]
Step-by-step explanation:
Given that:
[tex]\dfrac{dy}{dx} = xy^2[/tex]
and x = 1 when y = 1
by integrating:
first distribute all x variables on one side and all y variables on the other side.
[tex]\dfrac{dy}{y^2} = xdx[/tex]
now integrate both sides
[tex]\displaystyle\int {\dfrac{1}{y^2}}\,dy=\displaystyle\int {x} \, dx\\ \dfrac{-1}{y}=\dfrac{x^2}{2}+c[/tex]
to find the value of c, we can use the conditions given to us that x = 1 when y = 1.
[tex]-\dfrac{1}{y} = \dfrac{x^2}{2}+c\\-\dfrac{1}{1} = \dfrac{1^2}{2}+c\\c = -1-\dfrac{1}{2}\\c = -\dfrac{3}{2}[/tex]
now that we have our c, we can plug it into our integrated expression.
[tex]-\dfrac{1}{y} = \dfrac{x^2}{2}-\dfrac{3}{2}[/tex]
now simply rearrange the equation to make y the subject.
[tex]-\dfrac{1}{y} = \dfrac{x^2}{2}-\dfrac{3}{2}\\-y = \dfrac{2}{x^2-3}\\y = \dfrac{2}{3-x^2}[/tex]
this is the equation of y
