Let y be the solution of the differential equ | vedclass

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

Question

(A) The given differential equation is $(1-x^{2}) \frac{dy}{dx} = xy + (x^{3}+2) \sqrt{1-x^{2}}$.Rearranging,we get $\frac{dy}{dx} - \frac{x}{1-x^{2}}

Vedclass · Accepted Answer

$320$

Vedclass · Answer

(A) The given differential equation is $(1-x^{2}) \frac{dy}{dx} = xy + (x^{3}+2) \sqrt{1-x^{2}}$.Rearranging,we get $\frac{dy}{dx} - \frac{x}{1-x^{2}} y = \frac{x^{3}+2}{\sqrt{1-x^{2}}}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{-x}{1-x^{2}}$.The integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{-x}{1-x^{2}} dx} = e^{\frac{1}{2} \ln(1-x^{2})} = \sqrt{1-x^{2}}$.Multiplying by $IF$,we get $\frac{d}{dx} (y \sqrt{1-x^{2}}) = \frac{x^{3}+2}{\sqrt{1-x^{2}}} \cdot \sqrt{1-x^{2}} = x^{3}+2$.Integrating both sides,$y \sqrt{1-x^{2}} = \int (x^{3}+2) dx = \frac{x^{4}}{4} + 2x + C$.Given $y(0)=0$,we find $0 = 0 + 0 + C$,so $C=0$.Thus,$\sqrt{1-x^{2}} y(x) = \frac{x^{4}}{4} + 2x$.We need to evaluate $k = \int_{-\frac{1}{2}}^{\frac{1}{2}} (\frac{x^{4}}{4} + 2x) dx$.Since $2x$ is an odd function,$\int_{-\frac{1}{2}}^{\frac{1}{2}} 2x dx = 0$.Therefore,$k = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{x^{4}}{4} dx = 2 \int_{0}^{\frac{1}{2}} \frac{x^{4}}{4} dx = \frac{1}{2} [\frac{x^{5}}{5}]_{0}^{\frac{1}{2}} = \frac{1}{2} \cdot \frac{1}{5} \cdot \frac{1}{32} = \frac{1}{320}$.Thus,$k^{-1} = 320$.

Let $y$ be the solution of the differential equation $(1-x^{2}) dy = (xy + (x^{3}+2) \sqrt{1-x^{2}}) dx$ for $-1 < x < 1$ with $y(0)=0$. If $\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-x^{2}} y(x) dx = k$,then $k^{-1}$ is equal to:

Similar Questions

An integrating factor of the equation $(1+y+x^2 y) dx+(x+x^3) dy=0$ is

Let $y(x)$ be a solution of $(1+x^{2}) \frac{dy}{dx} + 2xy - 4x^{2} = 0$ and $y(0) = -1$. Then $y(1)$ is equal to

The general solution of the differential equation $\frac{d y}{d x}=\frac{2 y^2+1}{2 y^3-4 x y+y}$ is

The integrating factor of $x \frac{dy}{dx} + 3y = x^2$ is

Let $y=f(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^6+4x}{\sqrt{1-x^2}}$ for $-1 < x < 1$ such that $f(0)=0$. If $6 \int_{-1/2}^{1/2} f(x) dx = 2\pi - \alpha$,then $\alpha^2$ is equal to . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module