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

If $y = y(x)$,$x \in \left(0, \frac{\pi}{2}\right)$ is the solution curve of the differential equation $(\sin^2 2x) \frac{dy}{dx} + (8 \sin^2 2x + 2 \sin 4x) y = 2 e^{-4x} (2 \sin 2x + \cos 2x)$,with $y\left(\frac{\pi}{4}\right) = e^{-\pi}$,then $y\left(\frac{\pi}{6}\right)$ is equal to:

The solution of $(x+y+1) \frac{dy}{dx} = 1$ is

The differential equation $y^2 dx + (2xy - 1) dy = 0$ is

If $x=x(t)$ is the solution of the differential equation $(t+1) dx = (2x + (t+1)^4) dt$ with the initial condition $x(0) = 2$,then $x(1)$ equals:

The solution of the differential equation $\frac{dy}{dx} + \frac{3x^2}{1 + x^3}y = \frac{\sin^2 x}{1 + x^3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module