The solution of the differential equation fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

The solution of $\cos x \frac{dy}{dx} + y \sin x = 1$ is

Let $f:[1, \infty) \rightarrow [2, \infty)$ be a differentiable function such that $f(1)=2$. If $6 \int_1^x f(t) dt = 3x f(x) - x^3$ for all $x \geq 1$,then the value of $f(2)$ is

For the primitive integral equation $ydx + y^2dy = xdy$ ; $x \in R$,$y > 0$,$y = y(x)$,$y(1) = 1$,then $y(-3)$ is

Find the equation of a curve passing through the point $(0,2)$ given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by $5$.

$A$ function $y = f(x)$ satisfies $(x + 1)f'(x) - 2(x^2 + x)f(x) = \frac{e^{x^2}}{(x + 1)}$. If $f(0) = 5$,then $f(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module