1.

Find the equation of a curve, passes through `(-2,3)` at which the slope of tangent at any point `(x,y)` is `(2x)/(y^(2))`.

Answer» We know the slope of a curve at a point (x, y) is `(dy)/(dx)`.
`therefore (dy)/(dx)=(2x)/(y^(2)) " " ` …(i)
`rArr y^(2)dy = 2x dx " " `[separating the variables]
`rArr int y^(2)dy = int 2x dx `
`rArr (1)/(3)y^(3) = x^(2) +C " " `...(ii)
where C is a constant.
Thus, (ii) is the equation of the curve whose differential equation is given by (i) .
Since the given curve passes through the point (-2, 3), we have
`C=((1)/(3)xx 27)-(-2)^(2)=(9-4)=5.`
Hence , the required equation of the curve is
` (1)/(3) y^(3) = x^(2)+5 rArr y^(3) =3x^(2)+15`.


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