The solution of x dx + y dy = x^2 y dy - x y^ | vedclass

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(A) Given the differential equation: $x dx + y dy = x^2 y dy - x y^2 dx$ Rearranging the terms to group $x$ and $y$ variables: $x dx + x y^2 dx = x^2

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$x^2 - 1 = C(1 + y^2)$

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(A) Given the differential equation: $x dx + y dy = x^2 y dy - x y^2 dx$ Rearranging the terms to group $x$ and $y$ variables: $x dx + x y^2 dx = x^2 y dy - y dy$ $x(1 + y^2) dx = y(x^2 - 1) dy$ Separating the variables: $\frac{x}{x^2 - 1} dx = \frac{y}{1 + y^2} dy$ Multiplying both sides by $2$: $\frac{2x}{x^2 - 1} dx = \frac{2y}{1 + y^2} dy$ Integrating both sides: $\int \frac{2x}{x^2 - 1} dx = \int \frac{2y}{1 + y^2} dy$ $\ln|x^2 - 1| = \ln|1 + y^2| + \ln C$ Using the property $\ln a + \ln b = \ln(ab)$: $\ln|x^2 - 1| = \ln|C(1 + y^2)|$ Removing the logarithms: $x^2 - 1 = C(1 + y^2)$

$A \; (mm \; s^{-2})$	$16$	$8$	$0$	$-8$	$-16$
$x \; (mm)$	$-4$	$-2$	$0$	$2$	$4$

The solution of $x dx + y dy = x^2 y dy - x y^2 dx$ is

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