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

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

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$\frac{1}{x} = cy - y \log y$

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(B) Given the differential equation: $\frac{dx}{dy} + \frac{x}{y} = x^2$ Divide both sides by $x^2$: $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1$ Let $t = \frac{1}{x}$,then $\frac{dt}{dy} = -\frac{1}{x^2} \frac{dx}{dy}$ Substituting this into the equation: $-\frac{dt}{dy} + \frac{t}{y} = 1$ Rearranging gives: $\frac{dt}{dy} - \frac{t}{y} = -1$ This is a linear differential equation of the form $\frac{dt}{dy} + P(y)t = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = -1$ Integrating Factor ($I$.$F$.) $= e^{\int P(y) dy} = e^{-\int \frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$ The general solution is $t \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$ $t \cdot \frac{1}{y} = \int (-1) \cdot \frac{1}{y} dy + c$ $\frac{1}{xy} = -\log y + c$ Multiplying by $y$,we get: $\frac{1}{x} = cy - y \log y$

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

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