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 ;

${:\implies \quad \sf \dfrac{dy}{dx}=xy^{2}}$

Collect like terms in different sides ;

${:\implies \quad \sf \dfrac{dy}{y^2}=x\: dx}$

Integrating both sides will yield

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

${:\implies \quad \sf \dfrac{y^{-2+1}}{-2+1}=\dfrac{x^{1+1}}{1+1}+C}$

${:\implies \quad \sf \dfrac{-1}{y}=\dfrac{x^2}{2}+C}$

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 ;

${:\implies \quad \sf \dfrac{-1}{y}=\dfrac{x^2}{2}-\dfrac32}$

${:\implies \quad \sf -\dfrac{1}{y}=\dfrac{x^{2}-3}{2}}$

${:\implies \quad \sf -y=\dfrac{2}{x^{2}-3}}$

${:\implies \quad \bf \therefore \quad \underline{\underline{y=\dfrac{-2}{x^{2}-3}}}}$

Option B) is the required answer