I have to solve this equation, where $ a \in \mathbb R $:
$$ x y y' = x^2 + ( 1+ a ) y^2 $$
And I firstly thought about Bernouilli's method :
we set $ y^2 = z$ then the equation : $$ x \frac {z'} z = x^2 + (1+a) z $$
so then the homogenous equation gives : $$\frac{ z' } { z^2 } = \frac{1+a}{x} $$ so $$ \frac 1 z = \ln ( \frac{1}{ x^{1+a} })+ cst $$
It seems already false comparing to the answer I'm suppose to have. Can you tell me where is my mistake and if the substitution recommanded by Bernouilli's is a good idea?
You made a mistake with Bernouilli substitution..
$$x y y' = x^2 + ( 1+ a ) y^2$$ $$x\frac 12 (y^2)' = x^2 + ( 1+ a ) y^2$$ Then susbtitute $y^2=z$ $$x\frac 12 z' = x^2 + ( 1+ a )z$$ Which is linear of first order $$ z' - 2\frac {( 1+ a )}xz= 2x $$