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?

  • $\begingroup$ Write $\frac{dy}{y^2} = 3x^2 dx$ and integrate both sides. For the second, trying Googling "integrating factor". $\endgroup$
    – player3236
    Dec 17 '20 at 3:10
  • $\begingroup$ I understand the concept of applying the integrand to $e$, but the $y$ must stay behind in order for $5$ to become $5x$ $\endgroup$
    – DuncanK3
    Dec 18 '20 at 4:52
  • 1
    $\begingroup$ 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". $\endgroup$
    – player3236
    Dec 18 '20 at 4:57

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