The solution of frac{d y}{d x}+frac{1}{x}=fra | vedclass

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(A) Given differential equation is $\frac{d y}{d x}+\frac{1}{x}=\frac{e^y}{x^2}$.Dividing both sides by $e^y$,we get $e^{-y} \frac{d y}{d x} + \frac{1

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$2 x=\left(1+C x^2\right) e^y$

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(A) Given differential equation is $\frac{d y}{d x}+\frac{1}{x}=\frac{e^y}{x^2}$.Dividing both sides by $e^y$,we get $e^{-y} \frac{d y}{d x} + \frac{1}{x} e^{-y} = \frac{1}{x^2} \quad \dots (i)$.Let $e^{-y} = v$. Then,differentiating with respect to $x$,we get $-e^{-y} \frac{d y}{d x} = \frac{d v}{d x}$,or $e^{-y} \frac{d y}{d x} = -\frac{d v}{d x}$.Substituting this into equation $(i)$,we get $-\frac{d v}{d x} + \frac{1}{x} v = \frac{1}{x^2}$,which simplifies to $\frac{d v}{d x} - \frac{1}{x} v = -\frac{1}{x^2}$.This is a linear differential equation of the form $\frac{d v}{d x} + P(x) v = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = -\frac{1}{x^2}$.The integrating factor $IF = e^{\int P(x) d x} = e^{\int -\frac{1}{x} d x} = e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x}$.The solution is given by $v \cdot (IF) = \int Q(x) \cdot (IF) d x + C$.$v \cdot \frac{1}{x} = \int \left(-\frac{1}{x^2}\right) \cdot \frac{1}{x} d x + C$.$\frac{v}{x} = -\int x^{-3} d x + C = -\left(\frac{x^{-2}}{-2}\right) + C = \frac{1}{2 x^2} + C$.Multiplying by $2x^2$,we get $2 x v = 1 + 2 C x^2$. Let $2C = C'$,then $2 x v = 1 + C' x^2$.Substituting $v = e^{-y}$,we get $2 x e^{-y} = 1 + C' x^2$.Multiplying by $e^y$,we get $2 x = (1 + C' x^2) e^y$.

The solution of $\frac{d y}{d x}+\frac{1}{x}=\frac{e^y}{x^2}$ is

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