Let y=y(x) be the solution of the differentia | vedclass

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

Question

(C) The given differential equation is $(2x \ln x) \frac{dy}{dx} + 2y = \frac{3}{x} \ln x$.Dividing by $(2x \ln x)$,we get $\frac{dy}{dx} + \frac{y}{x

Vedclass · Accepted Answer

$-\frac{3}{e}$

Vedclass · Answer

(C) The given differential equation is $(2x \ln x) \frac{dy}{dx} + 2y = \frac{3}{x} \ln x$.Dividing by $(2x \ln x)$,we get $\frac{dy}{dx} + \frac{y}{x \ln x} = \frac{3}{2x^2}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \ln x}$ and $Q = \frac{3}{2x^2}$.The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \frac{1}{x \ln x} dx} = e^{\ln(\ln x)} = \ln x$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.$y \ln x = \int \frac{3}{2x^2} \ln x dx$.Using integration by parts,$\int \ln x \cdot (\frac{3}{2} x^{-2}) dx = \ln x \cdot (-\frac{3}{2x}) - \int \frac{1}{x} (-\frac{3}{2x}) dx = -\frac{3 \ln x}{2x} + \frac{3}{2} \int x^{-2} dx = -\frac{3 \ln x}{2x} - \frac{3}{2x} + C$.Given $y(e^{-1}) = 0$,so $0 \cdot \ln(e^{-1}) = -\frac{3 \ln(e^{-1})}{2e^{-1}} - \frac{3}{2e^{-1}} + C$.$0 = -\frac{3(-1)}{2e^{-1}} - \frac{3}{2e^{-1}} + C \Rightarrow 0 = \frac{3e}{2} - \frac{3e}{2} + C \Rightarrow C = 0$.Thus,$y \ln x = -\frac{3 \ln x}{2x} - \frac{3}{2x} \Rightarrow y = -\frac{3}{2x} - \frac{3}{2x \ln x}$.At $x = e$,$y(e) = -\frac{3}{2e} - \frac{3}{2e \ln e} = -\frac{3}{2e} - \frac{3}{2e} = -\frac{6}{2e} = -\frac{3}{e}$.

Let $y=y(x)$ be the solution of the differential equation $(2x \ln x) \frac{dy}{dx} + 2y = \frac{3}{x} \ln x$,$x > 0$ and $y(e^{-1}) = 0$. Then,$y(e)$ is equal to

Similar Questions

The Integrating Factor of the differential equation $x \frac{dy}{dx} + 2y = x^2$ $(x \neq 0)$ is . . . . . . .

Find the general solution of the differential equation $y dx - (x + 2y^2) dy = 0$.

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 $5$.

The integrating factor of the first order differential equation $x^{2}(x^{2}-1) \frac{dy}{dx} + x(x^{2}+1)y = x^{2}-1$ is

If $x y^{\prime}+y-e^x=0$ and $y(a)=b$, then $\lim _{x \rightarrow 1} y(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module