Answer to Question #186587 in Differential Equations for hammad sarwar

Question

: Solve Ricatti’s DEq. dy/dx =-2-y-y² , y₁=2

Answer 1

Given Differential equation: $\frac{dy}{dx}$ = 2 - y - y² , y(1) = 2

since it is a separable differential equation, therefore, separating it we get,

$\implies$ $\frac{dy}{2 - y - y^2}$ = $dx$

taking integral both side,

$\implies$ $\int$ $\frac{dy}{2 - y - y^2}$ = $\int$ $dx$

$\implies$ $\int$ $\frac{dy}{2 - y - y^2}$ = $\int$ $dx$

by using partial fraction decomposition we can write,

$\frac{dy}{2 - y - y^2}$ = $dy$ ( $\frac{-1}{3(y-1)}$ + $\frac{1}{3(y+2)}$ ) = $\frac{-1}{3(y-1)}dy$ + $\frac{1}{3(y+2)}dy$

therefore,

$\implies$ $\int$( $\frac{-1}{3(y-1)}dy$ + $\frac{1}{3(y+2)}dy$ ) = $\int$ $dx$

$\implies$ $\int$ $\frac{-1}{3(y-1)}dy$ + $\int$ $\frac{1}{3(y+2)}dy$ = $\int$ $dx$

$\implies$ $-\frac{1}{3}\ln \left(y-1\right)+\frac{1}{3}\ln \left(y+2\right)=t+C$

since given y(1) = 2

$-\frac{1}{3}\ln \left(2-1\right)+\frac{1}{3}\ln \left(2+2\right)=1+C$

$\implies$ C = $\frac{2\ln \left(2\right)}{3}-1$

hence solution of given Ricatti’s DEq is given by,

$\implies$ $-\frac{1}{3}\ln \left(y-1\right)+\frac{1}{3}\ln \left(y+2\right)=t+\frac{2\ln \left(2\right)}{3}-1$