Question

# Solve differential equationdy/dx+xy=xy^2

First order differential equations

Solve differential equation $$\frac{dy}{dx}+xy=xy^2$$

2021-01-14

$$\frac{dy}{dx}+xy=xy^2$$
$$\Rightarrow \frac{dy}{dx}= xy^2-xy$$
$$\Rightarrow \frac{dy}{dx}= xy(y-1)$$
$$\Rightarrow \frac{dy}{y(y-1)}= xdx$$
$$\frac{1}{y(y-1)}= \frac{A}{y}+\frac{B}{y-1}$$
$$\Rightarrow1= A(y-1)+By$$
$$\Rightarrow1= (A+B)y-A$$
Comparing the coefficient of y and constant term on both side, we get
$$A+B=0,\ -A=1\Rightarrow A= -1,\ B=1$$
So, we write
$$\frac{1}{y(y−1)}= -\frac{1}{y}+\frac{1}{(y−1)}$$
So, we have
$$(-\frac{1}{y}+\frac{1}{(y−1)}) dy= xdx$$
Now, we will integrate this equation on both the side
$$\int (-\frac{1}{y}+\frac{1}{(y-1)})dy= \int xdx$$ $$\Rightarrow -\ln(y)+\ln(y-1)= x^{\frac{2}{2}+C}$$
$$\Rightarrow \frac{y-1}{y}= x^{\frac{2}{2}+C}$$
$$\Rightarrow \frac{y-1}{y}= e^{x^{\frac{2}{2}+C}}$$
$$\Rightarrow \frac{y-1}{y}= e^{x^{\frac{2}{2}}}e^{C}$$
$$\Rightarrow \frac{y-1}{y}= C_{1} e^{x^{\frac{2}{2}}}$$, $$C_1=e^C$$
where C and $$C_1$$ are arbitrary constant
$$\Rightarrow y-1= C_{1} e^{\frac{x^{\frac{2}{2}}}{y}}$$
$$\Rightarrow y= C_{1} e^{x^{\frac{2}{2}}y}+1$$
$$\Rightarrow y-C e^{x^{\frac{2}{2}}}y= 1$$
$$\Rightarrow y(1-C_{1} e ^{x^{\frac{2}{2}}})=1$$
$$\Rightarrow y= \frac{1}{1-C_{1} e^{x^{\frac{2}{2}}}}$$