The solution of the differential equation fra | vedclass

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(A) The given differential equation is $\frac{dy}{dx} + \frac{y}{x \log_{e} x} = \frac{1}{x}$.This is a linear differential equation of the form $\fra

Vedclass · Accepted Answer

$2y = \log_{e} x + \frac{1}{\log_{e} x}$

Vedclass · Answer

(A) The given differential equation is $\frac{dy}{dx} + \frac{y}{x \log_{e} x} = \frac{1}{x}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{x \log_{e} x}$ and $Q = \frac{1}{x}$.The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int \frac{1}{x \log_{e} x} dx}$.Let $u = \log_{e} x$, then $du = \frac{1}{x} dx$. Thus, $\int \frac{1}{x \log_{e} x} dx = \int \frac{1}{u} du = \log_{e} u = \log_{e}(\log_{e} x)$.Therefore, $IF = e^{\log_{e}(\log_{e} x)} = \log_{e} x$.The solution is $y \cdot IF = \int (Q \cdot IF) dx + C$.$y \log_{e} x = \int \frac{1}{x} \log_{e} x dx$.Let $v = \log_{e} x$, then $dv = \frac{1}{x} dx$. The integral becomes $\int v dv = \frac{v^2}{2} + C = \frac{(\log_{e} x)^2}{2} + C$.So, $y \log_{e} x = \frac{(\log_{e} x)^2}{2} + C$.Given $y = 1$ when $x = e$, we have $1 \cdot \log_{e} e = \frac{(\log_{e} e)^2}{2} + C$.$1 = \frac{1}{2} + C \implies C = \frac{1}{2}$.Substituting $C$ back, $y \log_{e} x = \frac{(\log_{e} x)^2}{2} + \frac{1}{2}$.Dividing by $\log_{e} x$ (or multiplying by $2$), we get $2y \log_{e} x = (\log_{e} x)^2 + 1$, which simplifies to $2y = \log_{e} x + \frac{1}{\log_{e} x}$.

The solution of the differential equation $\frac{dy}{dx} + \frac{y}{x \log_{e} x} = \frac{1}{x}$ under the condition $y = 1$ when $x = e$ is

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