Solving nonlinear first order differential equations
I have the differential equation
$$x^{2}y'{(x)} + 2xy{(x)} = y^{2}{(x)}$$
With initial condition $y{(1)} = 1$ and I want to solve this, by observation, we can see the LHS is $x^{2}y{(x)}$ but I am unfamiliar with tackling the RHS and was wondering where to go from here.
$$x^{2}y{(x)} = \int y^{2}{(x)} \ dx$$
$\endgroup$4 Answers
$\begingroup$This suggests the change of unknown $z=x^2y$, leading to an equation in separeted variables.
$\endgroup$ $\begingroup$First, write the ode as
$$ x^{2}y'{(x)} + 2xy{(x)} = y^{2}{(x)} \implies y'+2\frac{y}{x}=\frac{y^2}{x^2}. $$
Now, use the change of variables $y=x u$ in the above ode which yields
$$ xu'+3u = u^2 \implies \int \frac{du}{u^2-3u} = \int \frac{dx}{x}. $$
I think you can finish it now.
$\endgroup$ $\begingroup$In addition to Julián's answer, note that the LHS is precisely the derivative of $x^2 y(x)$, which readily leads to the change of variables that he pointed out. Thus, you have:
$$\frac{dz}{dx} = \left(\frac{z}{x^2}\right)^2,$$
which can now easily solved.
Cheers!
$\endgroup$ $\begingroup$Note that $$ x^{2}y'{(x)} + 2xy{(x)} = [x^2 y(x)]' $$ Then $$ [x^2 y(x)]' = y^{2}{(x)} $$ Putting $z(x)=x^2 y(x)$, i.e. $z^2(x)/x^4=y^2(x)$, we have $$ z'(x) = \frac{z^2(x)}{x^4} $$ separable variables equation.
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