The solution of the differential equation y , | vedclass

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Question

(A) Given the differential equation: $y \, dx - x \, dy + x y^2 \, dx = 0$. Divide the entire equation by $y^2$ (assuming $y 
eq 0$): $\frac{y \, dx

Vedclass · Accepted Answer

$2x + x^2 y = \lambda y$

Vedclass · Answer

(A) Given the differential equation: $y \, dx - x \, dy + x y^2 \, dx = 0$. Divide the entire equation by $y^2$ (assuming $y 
eq 0$): $\frac{y \, dx - x \, dy}{y^2} + x \, dx = 0$. We recognize that $\frac{y \, dx - x \, dy}{y^2} = d\left( \frac{x}{y} ight)$. Substituting this into the equation,we get: $d\left( \frac{x}{y} ight) + x \, dx = 0$. Integrating both sides: $\int d\left( \frac{x}{y} ight) = - \int x \, dx$. $\frac{x}{y} = - \frac{x^2}{2} + c$,where $c$ is the constant of integration. Multiply by $2y$: $2x = -x^2 y + 2cy$. Rearranging the terms: $2x + x^2 y = (2c)y$. Let $\lambda = 2c$,then the solution is $2x + x^2 y = \lambda y$.

The solution of the differential equation $y \, dx - x \, dy + x y^2 \, dx = 0$ is:

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