If the solution of frac{dy}{dx} - y log_{e} 0 | vedclass

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

Question

(D) Given the differential equation $\frac{dy}{dx} - y \log_{e} 0.5 = 0$. Rearranging the terms,we get $\frac{dy}{dx} = y \log_{e} 0.5$. Separating th

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Given the differential equation $\frac{dy}{dx} - y \log_{e} 0.5 = 0$. Rearranging the terms,we get $\frac{dy}{dx} = y \log_{e} 0.5$. Separating the variables,we have $\frac{dy}{y} = (\log_{e} 0.5) dx$. Integrating both sides,$\int \frac{dy}{y} = \int (\log_{e} 0.5) dx$,which gives $\ln y = (\log_{e} 0.5) x + C$. Using the initial condition $y(0) = 1$,we substitute $x = 0$ and $y = 1$: $\ln 1 = (\log_{e} 0.5)(0) + C$,so $0 = 0 + C$,which implies $C = 0$. Thus,$\ln y = (\log_{e} 0.5) x$. Taking the exponential of both sides,$y = e^{(\log_{e} 0.5) x} = (e^{\log_{e} 0.5})^x = (0.5)^x$. We are given that $y(x) \rightarrow k$ as $x \rightarrow \infty$. Therefore,$k = \lim_{x \rightarrow \infty} (0.5)^x$. Since $0.5 Hence,$k = 0$.

If the solution of $\frac{dy}{dx} - y \log_{e} 0.5 = 0$,$y(0) = 1$,and $y(x) \rightarrow k$,as $x \rightarrow \infty$ then $k =$

Similar Questions

The general solution of the differential equation $2x \frac{dy}{dx} - y = 3$ is a family of

The general solution of the differential equation $(2x-y)^2 dy - 2(2x-y)^2 dx - 2 dx = 0$ is

$A$ particular solution of $\frac{dy}{dx} = (x + 9y)^2$,when $x = 0, y = \frac{1}{27}$ is

The solution of the equation $\frac{dy}{dx} = y(e^x + 1)$ is

Find the particular solution of the differential equation $\frac{dx}{dy} = \frac{\sin y(1 + y \cot y)}{x \log(x^2 e)}$,given that $y(1) = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module