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

If $-\frac{\pi}{4} < x < \frac{\pi}{4},$ then the general solution of the differential equation $\cos^{2} x \cdot \frac{dy}{dx} - (\tan 2x) y = \cos^{4} x$ is

The integrating factor of the differential equation $\frac{dy}{dx} + \frac{1}{x}y = x^3 - 3$ is

Let $f : R \rightarrow R$ be a differentiable function such that $f^{\prime}(x)+f(x)=\int \limits_0^2 f(t) dt$. If $f(0)=e^{-2}$,then $2f(0)-f(2)$ is equal to $.........$.

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{5}{x(x^5+1)}y=\frac{(x^5+1)^2}{x^7}$,for $x > 0$. If $y(1)=2$,then $y(2)$ is equal to

Let $f(x)=x-1$ and $g(x)=e^x$ for $x \in R$. If $\frac{d y}{d x}=\left(e^{-2 \sqrt{x}} g(f(f(x)))-\frac{y}{\sqrt{x}}\right)$ and $y(0)=0$,then $y(1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module