For x in R,let y(x) be a solution of the diff | vedclass

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

Question

(B) The given differential equation is $(x^2-5) \frac{dy}{dx} - 2xy = -2x(x^2-5)^2$.Dividing by $(x^2-5)$,we get $\frac{dy}{dx} - \frac{2x}{x^2-5} y =

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) The given differential equation is $(x^2-5) \frac{dy}{dx} - 2xy = -2x(x^2-5)^2$.Dividing by $(x^2-5)$,we get $\frac{dy}{dx} - \frac{2x}{x^2-5} y = -2x(x^2-5)$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2x}{x^2-5}$ and $Q(x) = -2x(x^2-5)$.The integrating factor $IF = e^{\int P(x) dx} = e^{-\int \frac{2x}{x^2-5} dx} = e^{-\ln|x^2-5|} = \frac{1}{x^2-5}$.The solution is $y \cdot IF = \int Q(x) \cdot IF dx + c$.$y \cdot \frac{1}{x^2-5} = \int -2x(x^2-5) \cdot \frac{1}{x^2-5} dx + c$.$\frac{y}{x^2-5} = \int -2x dx + c = -x^2 + c$.Given $y(2) = 7$,we have $\frac{7}{2^2-5} = -2^2 + c \Rightarrow \frac{7}{-1} = -4 + c \Rightarrow c = -3$.So,$\frac{y}{x^2-5} = -x^2 - 3$,which means $y = -(x^2-5)(x^2+3) = -(x^4 - 2x^2 - 15) = -x^4 + 2x^2 + 15$.To find the maximum value,let $t = x^2$ where $t \ge 0$. Then $y = -t^2 + 2t + 15$.This is a downward parabola with vertex at $t = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1$.Since $t=1$ is in the domain $t \ge 0$,the maximum value is $y(1) = -(1)^2 + 2(1) + 15 = -1 + 2 + 15 = 16$.

For $x \in R$,let $y(x)$ be a solution of the differential equation $(x^2-5) \frac{dy}{dx} - 2xy = -2x(x^2-5)^2$ such that $y(2)=7$. Find the maximum value of $y(x)$.

Similar Questions

The general solution of the differential equation $(\sec x + \tan x) \frac{dy}{dx} + (\sec^2 x + \sec x \tan x) y = 1$ is

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

The integrating factor of the differential equation $(1+x^{2}) \frac{dy}{dx} + y = e^{\tan^{-1} x}$ is

If the solution of the differential equation $x y^{\prime}=y+x^2 \sin x$ subject to the condition $y(\pi)=0$ is $y=f(x)$ and $f(x)$ has an extreme value at $x=\alpha$, then

If $\int_{a}^{x} t y(t) dt = x^2 + y(x)$,then $y$ as a function of $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module