The solution of frac{dx}{dy} + frac{x}{y} = x | vedclass

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(B) Given equation: $\frac{dx}{dy} + \frac{x}{y} = x^2$ Divide both sides by $x^2$: $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1 \dots (i)$ Let $\f

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$\frac{1}{x} = cy - y \log y$,where $c$ is a constant of integration.

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(B) Given equation: $\frac{dx}{dy} + \frac{x}{y} = x^2$ Divide both sides by $x^2$: $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1 \dots (i)$ Let $\frac{1}{x} = t$. Then $-\frac{1}{x^2} \frac{dx}{dy} = \frac{dt}{dy}$,so $\frac{1}{x^2} \frac{dx}{dy} = -\frac{dt}{dy}$. Substituting into $(i)$: $-\frac{dt}{dy} + \frac{t}{y} = 1$,which simplifies to $\frac{dt}{dy} - \frac{t}{y} = -1$. This is a linear differential equation in $t$. The Integrating Factor ($I$.$F$.) is $e^{\int -\frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$. The solution is $t(I.F.) = \int (-1)(I.F.) dy + c$. $t(\frac{1}{y}) = \int -\frac{1}{y} dy + c = -\log y + c$. Substituting $t = \frac{1}{x}$: $\frac{1}{xy} = -\log y + c$. Therefore,$\frac{1}{x} = cy - y \log y$.

The solution of $\frac{dx}{dy} + \frac{x}{y} = x^2$ is

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