Question

For the following differential equation verify that the accompanying function is a solution:

Differential equation	Function
\[x\frac{dy}{dx} + y = y^2\]	\[y = \frac{a}{x + a}\]

Answer 1

We have,
\[y = \frac{a}{x + a}\]
\[ \Rightarrow xy + ay = a\]
\[ \Rightarrow xy = a\left( 1 - y \right)\]
\[ \Rightarrow \frac{xy}{1 - y} = a\]
\[ \Rightarrow \frac{1 - y}{xy} = \frac{1}{a} . . . . . \left( 1 \right)\]
given differential equation: \[x\frac{dy}{dx} + y = y^2\]
Differentiating both sides of (1) with respect to x, we get 
\[\frac{xy\left( 0 - \frac{dy}{dx} \right) - \left( 1 - y \right)\left( x\frac{dy}{dx} + y \right)}{\left( xy \right)^2} = 0\]
\[ \Rightarrow xy\left( - \frac{dy}{dx} \right) - \left( 1 - y \right)\left( x\frac{dy}{dx} + y \right) = 0\]
\[ \Rightarrow - xy\frac{dy}{dx} - x\frac{dy}{dx} - y + xy\frac{dy}{dx} + y^2 = 0\]
\[ \Rightarrow - x\frac{dy}{dx} - y + y^2 = 0\]
\[ \Rightarrow x\frac{dy}{dx} + y = y^2\]
Hence, the given function is the solution to the given differential equation.