The differential equation x^2(y+1) dx + y^2(x | vedclass

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(D) Given the differential equation: $x^2(y+1) dx + y^2(x-1) dy = 0$. Rearranging the terms to separate variables: $\frac{x^2}{x-1} dx + \frac{y^2}{y+

Vedclass · Accepted Answer

$(x+1)^2+(y-1)^2+2 \log [(x-1)(y+1)]=C$

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(D) Given the differential equation: $x^2(y+1) dx + y^2(x-1) dy = 0$. Rearranging the terms to separate variables: $\frac{x^2}{x-1} dx + \frac{y^2}{y+1} dy = 0$. Integrating both sides: $\int \frac{x^2}{x-1} dx + \int \frac{y^2}{y+1} dy = C'$. Using polynomial division: $\frac{x^2}{x-1} = x+1 + \frac{1}{x-1}$ and $\frac{y^2}{y+1} = y-1 + \frac{1}{y+1}$. Substituting these into the integral: $\int (x+1 + \frac{1}{x-1}) dx + \int (y-1 + \frac{1}{y+1}) dy = C'$. Integrating term by term: $(\frac{x^2}{2} + x + \log |x-1|) + (\frac{y^2}{2} - y + \log |y+1|) = C'$. Multiplying by $2$: $x^2 + 2x + y^2 - 2y + 2 \log |(x-1)(y+1)| = 2C'$. Completing the squares: $(x^2 + 2x + 1) + (y^2 - 2y + 1) + 2 \log |(x-1)(y+1)| = 2C' + 2$. $(x+1)^2 + (y-1)^2 + 2 \log |(x-1)(y+1)| = C$.

The differential equation $x^2(y+1) dx + y^2(x-1) dy = 0$ has the general solution given by (where $C$ is a constant of integration.)

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