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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

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-example-1

1 Answer

12 votes

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

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