The general solution of the equation frac{dy} | vedclass

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

Question

(A) The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x}$ and $Q(x) = \frac{e^x}{x}$.First,we find

Vedclass · Accepted Answer

$y = \frac{e^x + c}{x}$

Vedclass · Answer

(A) The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x}$ and $Q(x) = \frac{e^x}{x}$.First,we find the integrating factor $(IF)$:$IF = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.The general solution is given by $y \cdot (IF) = \int Q(x) \cdot (IF) dx + c$.Substituting the values:$y \cdot x = \int \frac{e^x}{x} \cdot x dx + c$$xy = \int e^x dx + c$$xy = e^x + c$$y = \frac{e^x + c}{x}$.

The general solution of the equation $\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x}e^x$ is

Similar Questions

Let $y=y(x)$ be the solution curve of the differential equation $(1+x^{2})dy+(y-\tan^{-1}x)dx=0$,with $y(0)=1$. Then the value of $y(1)$ is:

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

The general solution of the differential equation $\left(\frac{1}{x^2}+x\right) \frac{d y}{d x}+3 y=1$ is

The general solution of the differential equation,$y' + y\phi'(x) - \phi(x)\phi'(x) = 0$,where $\phi(x)$ is a known function,is: (where $c$ is an arbitrary constant)

If $\cos x \frac{dy}{dx} - y \sin x = 6x$ for $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{3}) = 0$,then $y(\frac{\pi}{6})$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module