Question

find the differential equations for formula y'+1/3y=e^xy^4​

Answer 1

Step-by-step explanation:

We have,

$\frac{dy}{dx} + \frac{1}{3} y = {e}^{x} y^{4}$

$\implies \frac{1}{ {y}^{4} } \frac{dy}{dx} + \frac{1}{3 {y}^{3} } = {e}^{x}$

$Let \frac{1}{3 {y}^{3} } = v \\ \implies - \frac{1}{ {y}^{4} } \frac{dy}{dx} = \frac{dv}{dx}$

$\implies - \frac{dv}{dx} + v = {e}^{x}$

$\implies \frac{dv}{dx} - v = - {e}^{x}$

$I.F. = {e}^{ \int( - 1)dx} = {e}^{ - x}$

So,

$v .{e}^{ - x} = \int {e}^{x}. {e}^{ - x} dx$

$\implies \: v .{e}^{ - x} = \int dx$

$\implies \: v .{e}^{ - x} = x + C$

$\implies \: \frac{1}{3 {y}^{3} } .{e}^{ - x} = x + C$