Question

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

Answer 1

`y = (2)/(3-x^2)`

Explanation:

Given that:

`(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.

`(dy)/(y^2) = xdx`

now integrate both sides

`\displaystyle\int {(1)/(y^2)}\,dy=\displaystyle\int {x} \, dx\\ (-1)/(y)=(x^2)/(2)+c`

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

`-(1)/(y) = (x^2)/(2)+c\\-(1)/(1) = (1^2)/(2)+c\\c = -1-(1)/(2)\\c = -(3)/(2)`

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

`-(1)/(y) = (x^2)/(2)-(3)/(2)`

now simply rearrange the equation to make y the subject.

`-(1)/(y) = (x^2)/(2)-(3)/(2)\\-y = (2)/(x^2-3)\\y = (2)/(3-x^2)`

this is the equation of y