For the differential equation y^2 dx + left( | vedclass

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(C) Given the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$.Dividing by $y^2 dy$,we get $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$.Th

Vedclass · Accepted Answer

$1 + \frac{1}{y} - \frac{e^{\frac{1}{y}}}{e}$

Vedclass · Answer

(C) Given the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$.Dividing by $y^2 dy$,we get $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{y^2}$ and $Q(y) = \frac{1}{y^3}$.The Integrating Factor $(IF)$ is $e^{\int P(y) dy} = e^{\int \frac{1}{y^2} dy} = e^{-\frac{1}{y}}$.The general solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.$x e^{-\frac{1}{y}} = \int \frac{1}{y^3} e^{-\frac{1}{y}} dy + C$.Let $t = -\frac{1}{y}$,then $dt = \frac{1}{y^2} dy$. Also,$\frac{1}{y} = -t$.Substituting these,$x e^{-\frac{1}{y}} = \int (-t) e^t dt + C = - (t e^t - e^t) + C = (1 - t) e^t + C$.$x e^{-\frac{1}{y}} = (1 + \frac{1}{y}) e^{-\frac{1}{y}} + C$.Using $y(1) = 1$,we have $1 \cdot e^{-1} = (1 + 1) e^{-1} + C \Rightarrow e^{-1} = 2e^{-1} + C \Rightarrow C = -e^{-1} = -\frac{1}{e}$.Thus,$x e^{-\frac{1}{y}} = (1 + \frac{1}{y}) e^{-\frac{1}{y}} - \frac{1}{e}$.Dividing by $e^{-\frac{1}{y}}$,we get $x = 1 + \frac{1}{y} - \frac{e^{\frac{1}{y}}}{e}$.

For the differential equation $y^2 dx + \left( x - \frac{1}{y} \right) dy = 0$ with the initial condition $y(1) = 1$,find $x$.

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