Answer to Question #223126 in Differential Equations for Vakel

Question

Express 1 / y(y+1) in partial fractions.

Hence solve the differential equation dy/dx = y(y+1) / x given that y=4 when x=2 expressing y explicitly in terms of x.

Answer 1

Solution:

$\\ \dfrac1{y(y+1)}=\dfrac{A}{y}+\dfrac{B}{y+1} \\\Rightarrow \dfrac1{y(y+1)}=\dfrac{A(y+1)+By}{y(y+1)} \\\Rightarrow 1=Ay+A+By \\\Rightarrow 0.y+1=y(A+B)=A$

On comparing both sides,

$A+B=0,A=1 \\\Rightarrow B=-1$

Thus, $\dfrac1{y(y+1)}=\dfrac{1}{y}-\dfrac{1}{y+1}$ ...(i)

Now, given $\dfrac{dy}{dx}=\dfrac{y(y+1)}{x}$

$\Rightarrow \dfrac{1}{y(y+1)}dy=\dfrac{1}{x}dx \\ \Rightarrow [\dfrac{1}{y}-\dfrac{1}{y+1}]dy=\dfrac{1}{x}dx \ [using\ (i)]$

On integrating both sides,

$\ln |y|-\ln|y+1|=\ln |x|+\ln C \\\Rightarrow \ln|\dfrac{y}{y+1}|=\ln |\dfrac xC| \\\Rightarrow \dfrac{y}{y+1}=\dfrac xC$

Now, put y=4, x=2

$\dfrac{4}{5}=\dfrac 2C \\\Rightarrow C=\dfrac52$

Thus, the solution is $\dfrac{y}{y+1}=\dfrac {2x}{5}$