Question

Solve the differential equation ${\displaystyle \frac{dy}{dx}+xy=x{y}^2}$ and find the particular solution when y = 4, x = 1.

Answer 1

${\displaystyle \frac{dy}{dx}+xy=x{y}^2}$ 

∴ ${\displaystyle \frac{dy}{dx}=xy(y-1)}$

∴ ${\displaystyle \frac{dy}{y(y-1)}=x dx}$

∴ ${\displaystyle \int \frac{dy}{y(y-1)}=\int x dx}$

Let ${\displaystyle \frac{1}{y(y-1)}=\frac{A}{y}+\frac{B}{y-1}}$

∴ 1 = A(y – 1) + By

Put y = 1,

${\displaystyle \Rightarrow }$ B = 1,

Put y = 0,

${\displaystyle \Rightarrow }$ A = – 1

∴ ${\displaystyle \int (\frac{-1}{y}+\frac{1}{y-1})dy=\int x dx}$

∴  – log y + log(y – 1) = ${\displaystyle \frac{x^2}{2}+c}$

∴ ${\displaystyle \log \left(\frac{y-1}{y}\right)=\frac{x^2}{2}+c}$  ...(1)

Put y = 4,

x = 1

∴ ${\displaystyle \log \left(\frac{3}{4}\right)=\frac{1}{2}+c}$

∴ c = ${\displaystyle \log \left(\frac{3}{4}\right)-\frac{1}{2}}$

∴ From (1), the particular solution is

${\displaystyle \log \left(\frac{y-1}{y}\right)=\frac{x^2}{2}+\log \left(\frac{3}{4}\right)-\frac{1}{2}}$

i.e. ${\displaystyle \log \left[\frac{4(y-1)}{3y}\right]=\frac{x^2-1}{2}}$