Question

If dy / dx = xy2 and x = 1 when y = 1, then y = ?

Answer 1

Answer:

$y = \dfrac{2}{3-x^2}$

Step-by-step explanation:

Given that:

$\dfrac{dy}{dx} = xy^2$

and x = 1 when y = 1

by integrating:

first distribute all x variables on one side and all y variables on the other side.

$\dfrac{dy}{y^2} = xdx$

now integrate both sides

$\displaystyle\int {\dfrac{1}{y^2}}\,dy=\displaystyle\int {x} \, dx\\ \dfrac{-1}{y}=\dfrac{x^2}{2}+c$

to find the value of c, we can use the conditions given to us that x = 1 when y = 1.

$-\dfrac{1}{y} = \dfrac{x^2}{2}+c\\-\dfrac{1}{1} = \dfrac{1^2}{2}+c\\c = -1-\dfrac{1}{2}\\c = -\dfrac{3}{2}$

now that we have our c, we can plug it into our integrated expression.

$-\dfrac{1}{y} = \dfrac{x^2}{2}-\dfrac{3}{2}$

now simply rearrange the equation to make y the subject.

$-\dfrac{1}{y} = \dfrac{x^2}{2}-\dfrac{3}{2}\\-y = \dfrac{2}{x^2-3}\\y = \dfrac{2}{3-x^2}$

this is the equation of y