Answer to Question #306357 in Differential Equations for Francis

Question

ydx + (1-3y)xdy = 3y²e³ydy

Answer 1

ydx + (1-3y)xdy = 3y²$e^{3y}$ dy

=> $\frac{dx}{dy} + (\frac{1}{y}-3)x = 3ye^{3y}$

This is a linear differential equation of first order

Now integrating factor is given by,

I.F. = $e^{\int (\frac{1}{y}-3)dy}=e^{\ln y - 3y} = ye^{-3y}$

Multiplying both sides by integrating factor,

$ye^{-3y}$ $[\frac{dx}{dy} + (\frac{1}{y}-3)x]=$ $ye^{-3y}3ye^{3y}$

=> d{x$ye^{-3y}$ } = 3y²dy

=> $\int{d(xye^{-3y})}$ = $\int{3y²}dy$

=> xye$^{-3y}$ = $\frac{3y³}{3} + C$

=> xy $=e^{3y}(y³ + C)$ , $C$ is arbitrary constant.

This is the general solution of the given differential equation.