The function y=f(x) is the solution of the di | vedclass

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

Question

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{x}{x^2-1}$ and $Q

Vedclass · Accepted Answer

$\frac{\pi}{3}-\frac{\sqrt{3}}{4}$

Vedclass · Answer

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{x}{x^2-1}$ and $Q(x) = \frac{x^4+2x}{\sqrt{1-x^2}}$.Integrating factor ($I$.$F$.) $= e^{\int \frac{x}{x^2-1} dx} = e^{\frac{1}{2} \ln|x^2-1|} = e^{\frac{1}{2} \ln(1-x^2)} = \sqrt{1-x^2}$.The general solution is $y \cdot \sqrt{1-x^2} = \int \frac{x^4+2x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx + c = \int (x^4+2x) dx + c = \frac{x^5}{5} + x^2 + c$.Given $f(0)=0$,we have $0 \cdot 1 = 0 + 0 + c$,so $c=0$.Thus,$f(x) = \frac{x^5/5 + x^2}{\sqrt{1-x^2}}$.We need to evaluate $I = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$. Since $f(x) = \frac{x^5/5}{\sqrt{1-x^2}} + \frac{x^2}{\sqrt{1-x^2}}$,the first part is an odd function,so its integral over the symmetric interval is $0$.Thus,$I = 2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$.Let $x = \sin 	heta$,then $dx = \cos 	heta d	heta$. When $x=0, 	heta=0$; when $x=\frac{\sqrt{3}}{2}, 	heta=\frac{\pi}{3}$.$I = 2 \int_{0}^{\frac{\pi}{3}} \frac{\sin^2 	heta}{\cos 	heta} \cos 	heta d	heta = 2 \int_{0}^{\frac{\pi}{3}} \sin^2 	heta d	heta = \int_{0}^{\frac{\pi}{3}} (1 - \cos 2	heta) d	heta$.$I = [	heta - \frac{1}{2} \sin 2	heta]_{0}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{1}{2} \sin(\frac{2\pi}{3}) = \frac{\pi}{3} - \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$.

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

Similar Questions

The Integrating Factor of the differential equation $x \frac{dy}{dx} + 2y = x^2$ $(x \neq 0)$ is . . . . . . .

Let $y=y(x)$ be the solution of the differential equation $\cos x(3 \sin x+\cos x+3) dy = (1+y \sin x(3 \sin x+\cos x+3)) dx$; $0 \leq x \leq \frac{\pi}{2}, y(0)=0$. Then,$y\left(\frac{\pi}{3}\right)$ is equal to ..... .

If $y=y(x)$ is a particular solution of $\sqrt{1-x^2} \frac{dy}{dx} + \frac{2x}{\sqrt{1-x^2}} y = x$ with $y(0)=1$,then $y\left(\frac{1}{2}\right) = $

Let $y=y(x)$ be the solution of the differential equation $\sin x \frac{dy}{dx}+y \cos x=4x, x \in(0, \pi)$. If $y\left(\frac{\pi}{2}\right)=0$,then $y\left(\frac{\pi}{6}\right)$ is equal to

If $\frac{dy}{dx} + \frac{3}{\cos^2 x} y = \frac{1}{\cos^2 x}$,$x \in \left( -\frac{\pi}{3}, \frac{\pi}{3} \right)$ and $y\left( \frac{\pi}{4} \right) = \frac{4}{3}$,then $y\left( -\frac{\pi}{4} \right)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module