Question

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

Your answer:
0
x2
o
— 2
2 3
Q
x2 - 3
22 +1
2–3
2

Answer 1

Let's first start with the given that we have ;

`{:\implies \quad \sf (dy)/(dx)=xy^(2)}`

Collect like terms in different sides ;

`{:\implies \quad \sf (dy)/(y^2)=x\: dx}`

Integrating both sides will yield

`{:\implies \quad \displaystyle \sf \int y^(-2)dy=\int x^(1)dx}`

`{:\implies \quad \sf (y^(-2+1))/(-2+1)=(x^(1+1))/(1+1)+C}`

`{:\implies \quad \sf (-1)/(y)=(x^2)/(2)+C}`

Where, C is the Arbitrary Constant;

Now, as we are given that x = 1, when y = 1, so putting these values we will obtain C = (-3/2) , putting the values ;

`{:\implies \quad \sf (-1)/(y)=(x^2)/(2)-\frac32}`

`{:\implies \quad \sf -(1)/(y)=(x^(2)-3)/(2)}`

`{:\implies \quad \sf -y=(2)/(x^(2)-3)}`

`{:\implies \quad \bf \therefore \quad \underline{\underline{y=(-2)/(x^(2)-3)}}}`

Option B) is the required answer