# Initial value problems with y values

So I'm having trouble understanding how to do initial value problems using y values with respect to x. For instance,

$$\frac{dy}{dx} = 3x^{2}y^2$$ where $$y(1) = 1$$

I know you can divide both sides by $$y^2$$ and then integrating on the RHS will give $$x^3 + c$$ but how does the LHS operate?

Also, in a situation where $$\frac{dy}{dx} = 2 - 5y$$ where $$y(0) = 1$$ you can rearrange it to get $$\frac{dy}{dx} + 5y = 2$$ But then there is apparently an integrating factor of $$e^{5x}$$ Where does that come from? What steps am I missing?

• Write $\frac{dy}{y^2} = 3x^2 dx$ and integrate both sides. For the second, trying Googling "integrating factor". Dec 17 '20 at 3:10
• I understand the concept of applying the integrand to $e$, but the $y$ must stay behind in order for $5$ to become $5x$ Dec 18 '20 at 4:52
• Indeed, the method of integrating factors applies to ODEs of the form $y' + P(x) y = Q(x)$, producing the integrating factor $e^{\int P(x)dx}$. Hence the $y$ is "left behind". Dec 18 '20 at 4:57