If x=x(t) is the solution of the differential | vedclass

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

Question

(A) Given the differential equation: $(t+1) dx = (2x + (t+1)^4) dt$. Rearranging into the standard linear form $\frac{dx}{dt} + P(t)x = Q(t)$: $\frac{

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) Given the differential equation: $(t+1) dx = (2x + (t+1)^4) dt$. Rearranging into the standard linear form $\frac{dx}{dt} + P(t)x = Q(t)$: $\frac{dx}{dt} = \frac{2x + (t+1)^4}{t+1} = \frac{2x}{t+1} + (t+1)^3$. $\frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3$. The integrating factor $I.F. = e^{\int -\frac{2}{t+1} dt} = e^{-2 \ln(t+1)} = (t+1)^{-2} = \frac{1}{(t+1)^2}$. Multiplying both sides by $I.F.$: $\frac{d}{dt} \left( \frac{x}{(t+1)^2} \right) = (t+1)^3 \cdot \frac{1}{(t+1)^2} = (t+1)$. Integrating both sides with respect to $t$: $\frac{x}{(t+1)^2} = \int (t+1) dt = \frac{(t+1)^2}{2} + C$. Using the initial condition $x(0) = 2$: $\frac{2}{(0+1)^2} = \frac{(0+1)^2}{2} + C \Rightarrow 2 = \frac{1}{2} + C \Rightarrow C = \frac{3}{2}$. Thus,the solution is $x = \frac{(t+1)^4}{2} + \frac{3}{2}(t+1)^2$. For $t=1$: $x(1) = \frac{(1+1)^4}{2} + \frac{3}{2}(1+1)^2 = \frac{16}{2} + \frac{3}{2}(4) = 8 + 6 = 14$.

If $x=x(t)$ is the solution of the differential equation $(t+1) dx = (2x + (t+1)^4) dt$ with the initial condition $x(0) = 2$,then $x(1)$ equals:

Similar Questions

The solution of the differential equation $x^2 \frac{dy}{dx} - xy = 1$ is...

Let $y^{\prime}(x) + y(x) g^{\prime}(x) = g(x) g^{\prime}(x)$,$y(0) = 0$,$x \in \mathbb{R}$,where $f^{\prime}(x)$ denotes $\frac{d f(x)}{d x}$ and $g(x)$ is a given non-constant differentiable function on $\mathbb{R}$ with $g(0) = g(2) = 0$. Then the value of $y(2)$ is

Let $f : (0, \infty) \to (2, 20)$ be a twice differentiable function such that $\lim_{x \to \infty} (f(x) + f'(x) + f''(x)) = \lim_{x \to \infty} g(x)$,where $\lim_{x \to \infty} g(x)$ exists and is equal to $5$. Then $\lim_{x \to \infty} (f(x) - g(x))$ is equal to:

The integrating factor of the differential equation $\frac{dy}{dx} + y \tan x = \sec x$ is . . . . . . .

The integrating factor of the differential equation $x \frac{dy}{dx} + 2y = x^2 \log x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module