Differential Equation Exercise 2
In this exercise, we are tasked with solving the differential equation
$$ x^2y'=xy+y^2 $$
This is a first-order differential equation.
It is neither linear nor separable in its current form.
To isolate y', divide both sides of the equation by x2:
$$ \frac{ x^2y' }{x^2} = \frac{ xy+y^2 }{x^2} $$
$$ y' = \frac{ xy }{x^2} + \frac{ y^2 }{x^2} $$
$$ y' = \frac{ y }{x} + \frac{ y^2 }{x^2} $$
At this stage, it becomes clear that the equation is a first-order homogeneous differential equation of the form y' = f(y/x).
We introduce the substitution t = y/x:
$$ t = \frac{y}{x} $$
From this, we can express y and its derivative with respect to x:
$$ y = t \cdot x $$
$$ y' = D_x[ t \cdot x ] = t'x + t $$
Substituting y = tx into the original equation gives:
$$ y' = \frac{ y }{x} + \frac{ y^2 }{x^2} $$
$$ y' = t + t^2 $$
Now substitute y' = t'x + t into this result:
$$ t'x + t = t + t^2 $$
Expressing t' as dt/dx yields:
$$ \frac{dt}{dx} \cdot x + t = t + t^2 $$
Eliminating +t from both sides simplifies the equation to:
$$ \frac{dt}{dx} \cdot x = t^2 $$
This can now be solved using the separation of variables method.
Separating t and x:
$$ \frac{1}{t^2} \cdot dt = \frac{1}{x} \cdot dx $$
Integrating both sides with respect to their respective variables gives:
$$ \int \frac{1}{t^2} \cdot dt = \int \frac{1}{x} \cdot dx $$
The right-hand side integrates to log(x) + c:
$$ \int \frac{1}{t^2} \cdot dt = \log(x) + c $$
The left-hand side integrates to -1/t:
$$ \int t^{-2} \cdot dt = \log(x) + c $$
$$ \frac{ t^{-1} } { -1 } = \log(x) + c $$
$$ - \frac{ 1 } { t } = \log(x) + c $$
Recalling that t = y/x, we have:
$$ - \frac{ 1 } { \frac{y}{x} } = \log(x) + c $$
$$ - \frac{x}{y} = \log(x) + c $$
Solving for y gives:
$$ y = - \frac{x}{\log(x) + c} $$
This represents the general solution of the differential equation.
