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Kajal tried to solve the differential equation 
(dy)/(dx)=-x^(2)y^(2). This is her work:

(dy)/(dx)=-x^(2)y^(2)
Step 1: 
quad int-y^(-2)dy=intx^(2)dx
Step 2:

y^(-1)=(x^(3))/(3)+C
Step 3:

y=(3)/(x^(3))+C
Is Kajal's work correct? If not, what is her mistake?
Choose 1 answer:
(A) Kajal's work is correct.
(B) Step 1 is incorrect. The separation of variables wasn't done correctly.
(C) Step 2 is incorrect. Kajal didn't integrate 
x^(2) correctly.
(D) Step 3 is incorrect. Kajal didn't take the reciprocal of 
(x^(3))/(3)+C correctly.

Kajal tried to solve the differential equation dydx=x2y2 \frac{d y}{d x}=-x^{2} y^{2} . This is her work:\newlinedydx=x2y2 \frac{d y}{d x}=-x^{2} y^{2} \newlineStep 11: y2dy=x2dx \quad \int-y^{-2} d y=\int x^{2} d x \newlineStep 22: y1=x33+C\quad y^{-1}=\frac{x^{3}}{3}+C \newlineStep 33: y=3x3+C\quad y=\frac{3}{x^{3}}+C \newlineIs Kajal's work correct? If not, what is her mistake?\newlineChoose 11 answer:\newline(A) Kajal's work is correct.\newline(B) Step 11 is incorrect. The separation of variables wasn't done correctly.\newline(C) Step 22 is incorrect. Kajal didn't integrate x2 x^{2} correctly.\newline(D) Step 33 is incorrect. Kajal didn't take the reciprocal of x33+C \frac{x^{3}}{3}+C correctly.

Full solution

Q. Kajal tried to solve the differential equation dydx=x2y2 \frac{d y}{d x}=-x^{2} y^{2} . This is her work:\newlinedydx=x2y2 \frac{d y}{d x}=-x^{2} y^{2} \newlineStep 11: y2dy=x2dx \quad \int-y^{-2} d y=\int x^{2} d x \newlineStep 22: y1=x33+C\quad y^{-1}=\frac{x^{3}}{3}+C \newlineStep 33: y=3x3+C\quad y=\frac{3}{x^{3}}+C \newlineIs Kajal's work correct? If not, what is her mistake?\newlineChoose 11 answer:\newline(A) Kajal's work is correct.\newline(B) Step 11 is incorrect. The separation of variables wasn't done correctly.\newline(C) Step 22 is incorrect. Kajal didn't integrate x2 x^{2} correctly.\newline(D) Step 33 is incorrect. Kajal didn't take the reciprocal of x33+C \frac{x^{3}}{3}+C correctly.
  1. Check Separation of Variables: Kajal is attempting to solve the differential equation by separating variables. Let's check if the separation of variables in Step 11 is done correctly. The original equation is:\newline(dydx)=x2y2(\frac{dy}{dx}) = -x^2 y^2\newlineTo separate the variables, we should divide both sides by y2y^2 and multiply both sides by dxdx to get:\newline1y2dy=x2dx\frac{1}{y^2} dy = -x^2 dx\newlineNow, we integrate both sides:\newline(1y2)dy=(x2)dx\int(\frac{1}{y^2}) dy = \int(-x^2) dx\newlineThis looks like what Kajal has written, so Step 11 seems to be correct.
  2. Integrate Both Sides: Now, let's perform the integration on both sides. For the left side, the integral of 1y2\frac{1}{y^2} with respect to yy is 1y-\frac{1}{y}. For the right side, the integral of x2-x^2 with respect to xx is x33-\frac{x^3}{3}. So we have:\newline1y=x33+C-\frac{1}{y} = -\frac{x^3}{3} + C\newlineKajal's integration result for y1y^{-1} is correct, but she missed the negative sign on the right side of the equation.

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