If y(x) is the solution of the differential e | vedclass

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

Question

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 2 + \frac{1}{x}$ and $Q

Vedclass · Accepted Answer

$y(x)$ is decreasing in $\left( \frac{1}{2}, 1 \right)$

Vedclass · Answer

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 2 + \frac{1}{x}$ and $Q(x) = e^{-2x}$.The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int (2 + \frac{1}{x}) dx} = e^{2x + \log_e x} = x e^{2x}$.The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y(x e^{2x}) = \int e^{-2x} \cdot (x e^{2x}) dx + C = \int x dx + C = \frac{x^2}{2} + C$.Given $y(1) = \frac{1}{2}e^{-2}$,we substitute $x=1$ and $y=\frac{1}{2}e^{-2}$:$\frac{1}{2}e^{-2} \cdot (1 \cdot e^2) = \frac{1^2}{2} + C \Rightarrow \frac{1}{2} = \frac{1}{2} + C \Rightarrow C = 0$.Thus,$y = \frac{x}{2}e^{-2x}$.To check the monotonicity,we find $\frac{dy}{dx} = \frac{1}{2} e^{-2x} + \frac{x}{2} (-2 e^{-2x}) = \frac{e^{-2x}}{2} (1 - 2x)$.For $x \in \left( \frac{1}{2}, 1 \right)$,$1 - 2x < 0$,so $\frac{dy}{dx} < 0$,which means $y(x)$ is decreasing in $\left( \frac{1}{2}, 1 \right)$.

If $y(x)$ is the solution of the differential equation $\frac{dy}{dx} + \left( \frac{2x + 1}{x} \right)y = e^{-2x}, x > 0$ where $y(1) = \frac{1}{2}e^{-2}$,then:

Similar Questions

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 2y = f(x)$,where $f(x) = \begin{cases} 1, & x \in [0, 1] \\ 0, & \text{otherwise} \end{cases}$. If $y(0) = 0$,then $y\left(\frac{3}{2}\right)$ is

If the solution curve $f(x, y)=0$ of the differential equation $(1+\log_e x) \frac{dx}{dy} - x \log_e x = e^y, x > 0$,passes through the points $(1,0)$ and $(\alpha, 2)$,then $\alpha^\alpha$ is equal to

Solve the differential equation $\left(\tan ^{-1} y-x\right) d y=\left(1+y^{2}\right) d x$.

The solution of the equation $(x-4y^3) \frac{dy}{dx}-y=0, (y>0)$ is

The solution of the differential equation $(x + 2y^3) \frac{dy}{dx} = y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module