Let y=y(x), x1,be the solution of the differ | vedclass

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

Question

(B) The given differential equation is $(x-1) \frac{d y}{d x}+2 x y=\frac{1}{x-1}$.Dividing by $(x-1)$,we get $\frac{d y}{d x}+\frac{2 x}{x-1} y=\frac

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) The given differential equation is $(x-1) \frac{d y}{d x}+2 x y=\frac{1}{x-1}$.Dividing by $(x-1)$,we get $\frac{d y}{d x}+\frac{2 x}{x-1} y=\frac{1}{(x-1)^2}$.This is a linear differential equation of the form $\frac{d y}{d x}+P(x)y=Q(x)$,where $P(x)=\frac{2x}{x-1}=2+\frac{2}{x-1}$ and $Q(x)=\frac{1}{(x-1)^2}$.The integrating factor $IF = e^{\int P(x) dx} = e^{\int (2+\frac{2}{x-1}) dx} = e^{2x+2\ln(x-1)} = e^{2x}(x-1)^2$.The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.$y \cdot e^{2x}(x-1)^2 = \int \frac{1}{(x-1)^2} \cdot e^{2x}(x-1)^2 dx + C = \int e^{2x} dx + C = \frac{e^{2x}}{2} + C$.So,$y = \frac{e^{2x}}{2(x-1)^2 e^{2x}} + \frac{C}{(x-1)^2 e^{2x}} = \frac{1}{2(x-1)^2} + \frac{C}{e^{2x}(x-1)^2} = \frac{e^{2x} + 2C}{2e^{2x}(x-1)^2}$.Given $y(2) = \frac{1+e^4}{2e^4}$,we have $\frac{e^4 + 2C}{2e^4(1)^2} = \frac{1+e^4}{2e^4}$,which implies $2C = 1$,so $C = 1/2$.Thus,$y(x) = \frac{e^{2x}+1}{2e^{2x}(x-1)^2}$.For $x=3$,$y(3) = \frac{e^6+1}{2e^6(3-1)^2} = \frac{e^6+1}{8e^6}$.Comparing with $\frac{e^{\alpha}+1}{\beta e^{\alpha}}$,we get $\alpha=6$ and $\beta=8$.Therefore,$\alpha+\beta = 6+8 = 14$.

Let $y=y(x), x>1$,be the solution of the differential equation $(x-1) \frac{d y}{d x}+2 x y=\frac{1}{x-1}$,with $y(2)=\frac{1+e^{4}}{2 e^{4}}$. If $y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}}$,then the value of $\alpha+\beta$ is equal to

Similar Questions

Find the equation of a curve passing through the point $(0,1)$. If the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the $x$ coordinate (abscissa) and the product of the $x$ coordinate and $y$ coordinate (ordinate) of that point.

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x$ such that $y(0)=\frac{5}{4}$. Then $12\left(y\left(\frac{\pi}{4}\right)-e^{-2}\right)$ is equal to . . . . . . .

If the solution curve $f(x, y)=0$ of the differential equation $(1+\log_e x) \frac{dx}{dy} - x \log_e x = e^y, x > 0$,passes through the points $(1,0)$ and $(\alpha, 2)$,then $\alpha^\alpha$ is equal to

If $\frac{dx}{dy} = \frac{1+x-y^2}{y}$ and $x(1) = 1$,then $5x(2)$ is equal to:

The integrating factor of the differential equation $\frac{dy}{dx} + \frac{y}{x} = x^2$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module