The equation of the curve passing through the | vedclass

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

Question

(B) Let the point on the curve be $(x, y)$. The slope of the tangent is $\frac{dy}{dx}$.According to the problem,the sum of the ordinate $(y)$ and abs

Vedclass · Accepted Answer

$y=4-x-2 e^x$

Vedclass · Answer

(B) Let the point on the curve be $(x, y)$. The slope of the tangent is $\frac{dy}{dx}$.According to the problem,the sum of the ordinate $(y)$ and abscissa $(x)$ exceeds the slope by $5$,so:$y + x = \frac{dy}{dx} + 5$Rearranging the terms,we get the linear differential equation:$\frac{dy}{dx} - y = x - 5$This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x - 5$.The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.The general solution is $y \cdot IF = \int Q \cdot IF dx + C$:$y e^{-x} = \int (x - 5) e^{-x} dx + C$Using integration by parts for $\int (x - 5) e^{-x} dx$:$= (x - 5)(-e^{-x}) - \int (1)(-e^{-x}) dx$$= -(x - 5)e^{-x} - e^{-x} + C$$= (-x + 5 - 1)e^{-x} + C = (4 - x)e^{-x} + C$So,$y e^{-x} = (4 - x)e^{-x} + C$,which implies $y = 4 - x + C e^x$.Given the curve passes through $(0, 2)$:$2 = 4 - 0 + C e^0 \implies 2 = 4 + C \implies C = -2$.Substituting $C = -2$ into the equation:$y = 4 - x - 2 e^x$.

The equation of the curve passing through the point $(0,2)$ given that the sum of the ordinate and abscissa of any point exceeds the slope of the tangent to the curve at that point by $5$ is

Similar Questions

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

The general solution of $\frac{dy}{dx} + y f^{\prime}(x) - f(x) f^{\prime}(x) = 0$, where $y \neq f(x)$, is

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{2}y}{2\cos^4 x - \cos 2x} = x e^{\tan^{-1}(\sqrt{2} \cot 2x)}$,$0 < x < \pi/2$ with $y(\pi/4) = \pi^2/32$. If $y(\pi/3) = \frac{\pi^2}{18} e^{-\tan^{-1}(\alpha)}$,then the value of $3\alpha^2$ is equal to

The solution of the differential equation $ydx - (x + 2y^2)dy = 0$ is $x = f(y)$. If $f(-1) = 1$,then $f(1)$ is equal to

If $x dy + (y + y^2 x) dx = 0$ and $y = 1$ at $x = 1$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module