If a curve passes through the origin and the | vedclass

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

Question

(D) Given that the curve passes through the origin,so $y(0) = 0$.The slope of the tangent is given by $\frac{dy}{dx} = \frac{x^{2}-4x+y+8}{x-2}$.We ca

Vedclass · Accepted Answer

$(5, 5)$

Vedclass · Answer

(D) Given that the curve passes through the origin,so $y(0) = 0$.The slope of the tangent is given by $\frac{dy}{dx} = \frac{x^{2}-4x+y+8}{x-2}$.We can rewrite the numerator as $(x-2)^{2} + y + 4$,so $\frac{dy}{dx} = \frac{(x-2)^{2} + y + 4}{x-2} = (x-2) + \frac{y}{x-2} + \frac{4}{x-2}$.Rearranging the terms,we get the linear differential equation: $\frac{dy}{dx} - \frac{y}{x-2} = (x-2) + \frac{4}{x-2}$.The Integrating Factor $(I.F.)$ is $e^{-\int \frac{1}{x-2} dx} = e^{-\ln|x-2|} = \frac{1}{x-2}$.Multiplying both sides by the $I.F.$,we get $\frac{d}{dx} \left( \frac{y}{x-2} \right) = \frac{1}{x-2} \left( (x-2) + \frac{4}{x-2} \right) = 1 + \frac{4}{(x-2)^{2}}$.Integrating both sides with respect to $x$: $\frac{y}{x-2} = x - \frac{4}{x-2} + C$.Using the condition $y(0) = 0$: $\frac{0}{-2} = 0 - \frac{4}{-2} + C \Rightarrow 0 = 2 + C \Rightarrow C = -2$.Thus,$\frac{y}{x-2} = x - \frac{4}{x-2} - 2$.Multiplying by $(x-2)$,we get $y = x(x-2) - 4 - 2(x-2) = x^{2} - 2x - 4 - 2x + 4 = x^{2} - 4x$.Checking the options: For $x = 5$,$y = 5^{2} - 4(5) = 25 - 20 = 5$. Thus,the curve passes through $(5, 5)$.

If a curve passes through the origin and the slope of the tangent to it at any point $(x, y)$ is $\frac{x^{2}-4x+y+8}{x-2}$,then this curve also passes through the point

Similar Questions

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

The integrating factor of the differential equation $x \frac{dy}{dx} + y \log x = x e^x \cdot x^{-1/2} \log x$ for $x > 0$ is:

Let $y=y(x), y>0$,be a solution curve of the differential equation $(1+x^2) dy = y(x-y) dx$. If $y(0)=1$ and $y(2\sqrt{2})=\beta$,then

The solution of the differential equation $\frac{dy}{dx} + x \sin^2 y = \sin y \cos y$ is:

If $u(x)$ and $v(x)$ are two independent solutions of the differential equation $\frac{d^{2} y}{d x^{2}}+b \frac{d y}{d x}+c y=0$, then which of the following is also a solution of the given differential equation?

Vedclass Test Series

Exam Paper Generator

Online Exam Module