Let f be a differentiable function with lim_{ | 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 $P(x) = f^{\prime}(x)$ and $Q(

Vedclass · Accepted Answer

$y + 1 = e^{-f(x)} + f(x)$

Vedclass · Answer

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = f^{\prime}(x)$ and $Q(x) = f(x)f^{\prime}(x)$.The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int f^{\prime}(x) dx} = e^{f(x)}$.The general solution is given by $y \cdot IF = \int Q(x) \cdot IF dx + C$.$y \cdot e^{f(x)} = \int f(x) f^{\prime}(x) e^{f(x)} dx + C$.Let $u = f(x)$, then $du = f^{\prime}(x) dx$. The integral becomes $\int u e^u du = u e^u - e^u$.Thus, $y \cdot e^{f(x)} = e^{f(x)}(f(x) - 1) + C$.Given $\lim_{x \rightarrow \infty} f(x) = 0$ and $\lim_{x \rightarrow \infty} y(x) = 0$, we substitute these into the equation:$0 \cdot e^0 = e^0(0 - 1) + C \Rightarrow 0 = -1 + C \Rightarrow C = 1$.Substituting $C = 1$ back into the equation:$y \cdot e^{f(x)} = e^{f(x)}(f(x) - 1) + 1$.Dividing by $e^{f(x)}$:$y = f(x) - 1 + e^{-f(x)}$.Rearranging gives $y + 1 = e^{-f(x)} + f(x)$.

Let $f$ be a differentiable function with $\lim_{x \rightarrow \infty} f(x) = 0$. If $y^{\prime} + y f^{\prime}(x) - f(x) f^{\prime}(x) = 0$ and $\lim_{x \rightarrow \infty} y(x) = 0$, then (where $y^{\prime} = \frac{dy}{dx}$):

Similar Questions

The solution of the differential equation $(1+x) \frac{dy}{dx} - xy = 1-x$ is

Let $f:[1, \infty) \rightarrow \mathbb{R}$ be a differentiable function. If $6 \int_{1}^{x} f(t) dt = 3xf(x) + x^{3} - 4$ for all $x \ge 1$, then the value of $f(2) - f(3)$ is

The solution of the differential equation $ydx - (x + 2y^2)dy = 0$ is $x = f(y)$. If $f(-1) = 1$,then $f(1)$ is equal to

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module