The general solution of the differential equa | vedclass

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(C) Given differential equation is $\frac{dx}{dy} + \frac{x}{y} = x^2$. Dividing both sides by $x^2$,we get $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy

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

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(C) Given differential equation is $\frac{dx}{dy} + \frac{x}{y} = x^2$. Dividing both sides by $x^2$,we get $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1$. Let $t = \frac{1}{x}$,then $\frac{dt}{dy} = -\frac{1}{x^2} \frac{dx}{dy}$. Substituting this into the equation,we get $-\frac{dt}{dy} + \frac{t}{y} = 1$,which simplifies to $\frac{dt}{dy} - \frac{t}{y} = -1$. This is a linear differential equation of the form $\frac{dt}{dy} + P(y)t = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = -1$. The integrating factor $I.F. = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$. The solution is $t \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$. $t \cdot \frac{1}{y} = \int (-1) \cdot \frac{1}{y} dy + c$. $\frac{t}{y} = -\log |y| + c$. Substituting $t = \frac{1}{x}$,we get $\frac{1}{xy} = -\log |y| + c$,or $\frac{1}{x} = cy - y \log |y|$.

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

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