The solution of x frac{dy}{dx} = y + x e^{y/x | vedclass

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

Question

(D) Given differential equation is $x \frac{dy}{dx} = y + x e^{y/x}$.Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} + e^{y/x}$.This is a homogene

Vedclass · Accepted Answer

$e^{-y/x} + \log x = 1$

Vedclass · Answer

(D) Given differential equation is $x \frac{dy}{dx} = y + x e^{y/x}$.Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} + e^{y/x}$.This is a homogeneous differential equation.Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting this into the equation: $v + x \frac{dv}{dx} = v + e^v$.This simplifies to $x \frac{dv}{dx} = e^v$.Separating variables: $e^{-v} dv = \frac{1}{x} dx$.Integrating both sides: $\int e^{-v} dv = \int \frac{1}{x} dx$,which gives $-e^{-v} = \log x + c$.Substituting $v = y/x$: $-e^{-y/x} = \log x + c$.Using the condition $y(1) = 0$: $-e^{-0/1} = \log 1 + c \Rightarrow -1 = 0 + c \Rightarrow c = -1$.Thus,$-e^{-y/x} = \log x - 1$,which rearranges to $e^{-y/x} + \log x = 1$.

The solution of $x \frac{dy}{dx} = y + x e^{y/x}$ with $y(1) = 0$ is

Similar Questions

The potential energy between two atoms in a molecule is given by $U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}$,where $a$ and $b$ are positive constants and $x$ is the distance between the atoms. The atoms are in stable equilibrium when:

In printed scientific names, only the . . . . . . is capitalized.

The moment of inertia of a meter scale of mass $0.6\, kg$ about an axis perpendicular to the scale and located at the $20\, cm$ position on the scale in $kg\, m^2$ is (Breadth of the scale is negligible).

What is the concentration of $20$ volume $H_2O_2$ in $g/L$?

Heating an aqueous solution of aluminium chloride to dryness will give:

Vedclass Test Series

Exam Paper Generator

Online Exam Module