How could I solve this IVP?

Question

$y'+y\cdot \ln^2(x)=y^2\cdot \ln^2(x)$

I tried transforming it to $y'+P(x)y=Q(x)$ but I'm not sure how

Answer 1

Hint: $$y’=(\ln x)^2 (y^2-y)=(\ln x)^2\bigg(\left(y-\frac 12\right)^2 -\frac 14 \bigg)$$

Just separate the variables now.

Answer 2

$\textbf{HINT}$ Set $z=1/y$, so that the differential equation transforms into

$$-z'+z\cdot\ln^2 x =\ln^2 x$$

Answer 3

$$y'+y\cdot \ln^2(x)=y^2\cdot \ln^2(x)$$ It's Bernoulli's differential equation. Substitute $u=\dfrac 1 y$ then it becomes linear of first order: $$-u'+u \ln^2x=\ln^2 x$$

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The differential equation is also separable: $$ \int \dfrac {dy}{y(y-1)}=\int\ln^2(x)dx$$ $$ \int \dfrac {dy}{y-1}-\int \dfrac {dy}{y}=\int\ln^2(x)dx$$ Evaluate the last integral by integration by part: $$I=\int\ln^2(x)dx$$ $$I=x \ln^2 x -2x( \ln x -1)+C$$