Let the solution curve y = y(x) of the differ | vedclass

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

Question

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{3x^5 	an^{-1}(x

Vedclass · Accepted Answer

$\exp \left( \frac{4-\pi}{4 \sqrt{2}} \right)$

Vedclass · Answer

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{3x^5 	an^{-1}(x^3)}{(1+x^6)^{3/2}}$ and $Q(x) = 2x \exp \left( \frac{x^3 - 	an^{-1}(x^3)}{\sqrt{1+x^6}} ight)$.First,we find the Integrating Factor $(I.F.)$:$I.F. = e^{\int P(x) dx} = e^{\int -\frac{3x^5 	an^{-1}(x^3)}{(1+x^6)^{3/2}} dx}$.Let $t = x^3$,then $dt = 3x^2 dx$. The integral becomes $\int -\frac{t 	an^{-1}(t)}{(1+t^2)^{3/2}} dt$. Using integration by parts,we get $\frac{	an^{-1}(t) - t}{\sqrt{1+t^2}} = \frac{	an^{-1}(x^3) - x^3}{\sqrt{1+x^6}}$.Thus,$I.F. = \exp \left( \frac{	an^{-1}(x^3) - x^3}{\sqrt{1+x^6}} ight)$.The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y \cdot \exp \left( \frac{	an^{-1}(x^3) - x^3}{\sqrt{1+x^6}} ight) = \int 2x \exp \left( \frac{x^3 - 	an^{-1}(x^3)}{\sqrt{1+x^6}} ight) \cdot \exp \left( \frac{	an^{-1}(x^3) - x^3}{\sqrt{1+x^6}} ight) dx + C$.$y \cdot \exp \left( \frac{	an^{-1}(x^3) - x^3}{\sqrt{1+x^6}} ight) = \int 2x dx + C = x^2 + C$.Since the curve passes through the origin $(0,0)$,we have $0 \cdot e^0 = 0^2 + C$,so $C = 0$.Thus,$y(x) = x^2 \exp \left( \frac{x^3 - 	an^{-1}(x^3)}{\sqrt{1+x^6}} ight)$.At $x = 1$,$y(1) = 1^2 \exp \left( \frac{1 - 	an^{-1}(1)}{\sqrt{1+1}} ight) = \exp \left( \frac{1 - \pi/4}{\sqrt{2}} ight) = \exp \left( \frac{4-\pi}{4 \sqrt{2}} ight)$.

Let the solution curve $y = y(x)$ of the differential equation $\frac{dy}{dx} - \frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}} y = 2x \exp \left( \frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}} \right)$ pass through the origin. Then $y(1)$ is equal to:

Similar Questions

Let $f: R \rightarrow R$ be a continuous function satisfying $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0$ for all $x \in R$. Then:

Let $y=y(x)$ be a solution of the differential equation $(x \cos x) dy + (xy \sin x + y \cos x - 1) dx = 0$,$0 < x < \frac{\pi}{2}$. If $\frac{\pi}{3} y(\frac{\pi}{3}) = \sqrt{3}$,then $|\frac{\pi}{6} y''(\frac{\pi}{6}) + 2 y'(\frac{\pi}{6})|$ is equal to $.........$.

If $y=\sin x+A \cos x$ is the general solution of $\frac{dy}{dx}+f(x)y=\sec x$, then an integrating factor of the differential equation is

The function $y=f(x)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}}$ in $(-1,1)$ satisfying $f(0)=0$. Then $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$ is

If $x \log x \frac{dy}{dx} + y = \log x^2$ and $y(e) = 0$,then $y(e^2) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module