Question

dy/dx=e^3x + 2y
plz help me to solve this​

Answer 1

Answer:

which formula we should use bro tell me fast

Answer 2

Working Note :-

Linear Differential equation of the form,

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

We have to first find the Integrating Factor, I.F.,

$\rm :\longmapsto\:I.F. = {e}^{ \: \int \: pdx}$

Then, solution of differential equation is given by

$\rm :\longmapsto\:y \times \: I.F. = \int \: q \times I.F.dx$

Let's solve the problem now!!!

Consider,

$\rm :\longmapsto\:\dfrac{dy}{dx} = 2y + {e}^{3x} \:$

$\rm :\longmapsto\:\dfrac{dy}{dx} - 2y = {e}^{3x} \:$

On comparing with,

$\red{\rm :\longmapsto\:\dfrac{dy}{dx} + py = q \: }$

we get

$\rm :\longmapsto\:\blue{ \bf \: p = - 2 \: and \: q = {e}^{3x}}$

Now,

$\rm :\longmapsto\:I.F. = {e}^{ \: \int \: pdx}$

$\rm :\longmapsto\:I.F. = {e}^{ \: \int \: - 2dx}$

$\rm :\longmapsto\:I.F. = {e}^{ - 2x}$

Hence,

Solution of differential equation is

$\rm :\longmapsto\:y \times \: I.F. = \int \: q \times I.F.dx$

$\rm :\longmapsto\:y \times \: {e}^{ - 2x} = \int {e}^{ - 2x} \times {e}^{3x} dx$

$\rm :\longmapsto\:y \times \: {e}^{ - 2x} = \int {e}^{ - 2x + 3x}dx$

$\rm :\longmapsto\:y \times \: {e}^{ - 2x} = \int {e}^{x}dx$

$\rm :\longmapsto\:y \times \: {e}^{ - 2x} = {e}^{x} \: + \: c$