Let y=y(x) be the solution of the differentia | 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 $P(x) = \frac{5}{x(x^5+1)}$ and

Vedclass · Accepted Answer

$\frac{693}{128}$

Vedclass · Answer

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{5}{x(x^5+1)}$ and $Q(x) = \frac{(x^5+1)^2}{x^7}$.The Integrating Factor ($I$.$F$.) is given by $e^{\int P(x) dx} = e^{\int \frac{5}{x(x^5+1)} dx}$.To solve the integral,multiply the numerator and denominator by $x^{-6}$:$I.F. = e^{\int \frac{5x^{-6}}{x^{-5}+1} dx}$.Let $t = x^{-5}+1$,then $dt = -5x^{-6} dx$,so $-dt = 5x^{-6} dx$.$I.F. = e^{\int \frac{-dt}{t}} = e^{-\ln|t|} = \frac{1}{t} = \frac{1}{x^{-5}+1} = \frac{x^5}{x^5+1}$.The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y \cdot \frac{x^5}{x^5+1} = \int \frac{(x^5+1)^2}{x^7} \cdot \frac{x^5}{x^5+1} dx + C = \int \frac{x^5+1}{x^2} dx + C = \int (x^3 + x^{-2}) dx + C$.$y \cdot \frac{x^5}{x^5+1} = \frac{x^4}{4} - \frac{1}{x} + C$.Given $y(1) = 2$,we have $2 \cdot \frac{1}{1+1} = \frac{1}{4} - 1 + C \Rightarrow 1 = -\frac{3}{4} + C \Rightarrow C = \frac{7}{4}$.Thus,$y \cdot \frac{x^5}{x^5+1} = \frac{x^4}{4} - \frac{1}{x} + \frac{7}{4}$.For $x=2$,$y \cdot \frac{32}{33} = \frac{16}{4} - \frac{1}{2} + \frac{7}{4} = 4 - 0.5 + 1.75 = 5.25 = \frac{21}{4}$.$y = \frac{21}{4} \cdot \frac{33}{32} = \frac{693}{128}$.

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

Similar Questions

The general solution of the differential equation $(y^2+x+1) dy = (y+1) dx$ is

Let $y=y(x)$ be the solution of the differential equation $2 \cos x \frac{d y}{d x}=\sin 2 x-4 y \sin x$,where $x \in \left(0, \frac{\pi}{2}\right)$. If $y\left(\frac{\pi}{3}\right)=0$,then $y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)$ is equal to:

If $\tan x$ is an integrating factor of the differential equation $\frac{dy}{dx} + Py = Q$, then $P$ can be

The general solution of the differential equation $(1+\tan y)(dx-dy)+2x dy=0$ is

If $f: R \rightarrow R$ is a differentiable function such that $f^{\prime}(x) > 2f(x)$ for all $x \in R$,and $f(0) = 1$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module