If the slope of the tangent drawn at any poin | vedclass

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

Question

(A) The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = x + y$. Rearranging the equation,we get $\frac{dy}{dx} - y = x$. This

Vedclass · Accepted Answer

$y=ce^x-x-1$

Vedclass · Answer

(A) The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = x + y$. Rearranging the equation,we get $\frac{dy}{dx} - y = x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x$. The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int -1 dx} = e^{-x}$. Multiplying both sides by the integrating factor,we get $e^{-x} \frac{dy}{dx} - e^{-x} y = x e^{-x}$. This can be written as $\frac{d}{dx} (y e^{-x}) = x e^{-x}$. Integrating both sides with respect to $x$,we get $y e^{-x} = \int x e^{-x} dx$. Using integration by parts,$\int x e^{-x} dx = x(-e^{-x}) - \int 1(-e^{-x}) dx = -x e^{-x} - e^{-x} + c$. Thus,$y e^{-x} = -x e^{-x} - e^{-x} + c$. Multiplying by $e^x$,we get $y = -x - 1 + ce^x$,which is $y = ce^x - x - 1$.

If the slope of the tangent drawn at any point $(x, y)$ on a curve is $(x+y)$,then the equation of that curve is

Similar Questions

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

Integrating factor of the differential equation $\frac{dy}{dx} + y \tan x = \sec x$ is

If $y'' - 3y' + 2y = 0$ where $y(0) = 1$ and $y'(0) = 0$, then the value of $y$ at $x = \log_{e} 2$ is

Let $y=y(t)$ be a solution of the differential equation $\frac{dy}{dt}+\alpha y=\gamma e^{-\beta t}$,where $\alpha > 0, \beta > 0$ and $\gamma > 0$. Then $\lim_{t \rightarrow \infty} y(t)$ is:

Find the general solution of the differential equation: $x \frac{dy}{dx} + 2y = x^2 \log x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module