Question

what is the solution of the following differentiatial equation ln(dy/dx) +y=x. ?​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Solve the differential equation :-

$\rm :\longmapsto\: log\bigg(\dfrac{dy}{dx}\bigg) + y = x$

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm :\longmapsto\: log\bigg(\dfrac{dy}{dx}\bigg) + y = x$

can be rewritten as

$\rm :\longmapsto\: log\bigg(\dfrac{dy}{dx}\bigg) = x - y$

We know,

$\red{\rm :\longmapsto\:If \: \: log(y) = x \: \: then \: \: y = {e}^{x}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {e}^{x - y}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {e}^{x} \times {e}^{ - y}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ {e}^{x} }{ {e}^{y} }$

Now, on separating the variables, we get

$\rm :\longmapsto\: {e}^{y}dy = {e}^{x}dx$

On integrating both sides, we get

$\rm :\longmapsto\: \displaystyle\int\rm {e}^{y}dy = \displaystyle\int\rm {e}^{x}dx$

$\rm :\longmapsto\: {e}^{y} = {e}^{x} + c$

Additional Information :-

The linear Differential equation is of the form

$\red{\rm :\longmapsto\:\dfrac{dy}{dx} + py = q \: \: where \: p, \: q \: \in \: f(x)}$

Then

Step :- 1 Integrating Factor

$\red{\rm :\longmapsto\:IF \: = \: {e} \: ^ {\displaystyle\int\rm pdx} }$

and

Step :- 2 Solution is given by

$\red{\rm :\longmapsto\:y \times IF = \displaystyle\int\rm (q \times IF )\: dx}$