Answer to Question #239133 in Differential Equations for Anuj

Question

Express 2/y²-1 in partial fractions. Hence solve the differential equation 2xdy/dx +1 = y² given that y= -3 when x= 1, expressing y explicitly in terms of x.

Answer 1

It follows that $\frac{2}{y^2-1}=\frac{1}{y-1}-\frac{1}{y+1}$ is the expression in partial fractions.

Let us solve the differential equation $\frac{2xdy}{dx} +1 = y^2,$  given that $y= -3$ when $x= 1.$ This equation is equvalent to the equation $\frac{2xdy}{dx} = y^2-1,$ and hencev $\frac{2dy}{y^2-1} = \frac{dx}x.$ It follows that $\int\frac{2dy}{y^2-1} = \int\frac{dx}x,$ and hence $\int(\frac{1}{y-1}-\frac{1}{y+1})dy= \ln|x|+\ln|C|.$ We conclude that $\ln|y-1|-\ln|y+1|=\ln|Cx|,$ that is $\ln|\frac{y-1}{y+1}|=\ln|Cx|,$ and hence $\frac{y-1}{y+1}=Cx.$ Taking into account that $y= -3$ when $x= 1,$ we conclude that $C=\frac{-3-1}{-3+1}=2.$

It follows that $y-1=2xy+2x,$ and hence $y-2xy=2x+1.$ Consequently, $y=\frac{2x+1}{1-2x}$ is the solution of the differential equation.