If the curve y = f(x) passes through the poin | vedclass

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

Question

(C) Given the differential equation: $\frac{dy}{y} = (2 + \ln x) dx$.Integrating both sides: $\int \frac{dy}{y} = \int (2 + \ln x) dx$.$\ln y = 2x + (

Vedclass · Accepted Answer

$e^{2e}$

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{y} = (2 + \ln x) dx$.Integrating both sides: $\int \frac{dy}{y} = \int (2 + \ln x) dx$.$\ln y = 2x + (x \ln x - x) + C = x \ln x + x + C$.Since the curve passes through $(1, e)$, substitute $x = 1$ and $y = e$: $\ln e = 1 \ln 1 + 1 + C$.$1 = 0 + 1 + C$, which gives $C = 0$.Thus, the equation of the curve is $\ln y = x \ln x + x$.To find $f(e)$, substitute $x = e$: $\ln f(e) = e \ln e + e = e(1) + e = 2e$.Therefore, $f(e) = e^{2e}$.

If the curve $y = f(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \log_e x) dx, x > 0$, then $f(e)$ is equal to:

Similar Questions

The general solution of the differential equation $e^{\frac{1}{2}\left(\frac{dy}{dx}\right)}=3^x$ is (where $C$ is a constant of integration).

If $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$,then $f(xy)$ is equal to

For solving $\frac{dy}{dx} = (4x + y + 1)$,the suitable substitution is:

The slope of a curve at any point is inversely proportional to twice the $y$-coordinate of that point. If the curve passes through $(4, 3)$,then the equation of the curve is:

The general solution of the differential equation $\tan x \tan y \, dx + \cos^2 x \operatorname{cosec}^2 y \, dy = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module