$\frac{dy}{dx} = \frac{(y+1)^4}{[1-3((y+1)^3)]x}, \; y(0)=0$

I have to solve it using an integrating factor. I subtracted the right hand side to the left hand side to make everything equal 0. I'm not sure what to do from there. Am I dealing with a Bernoulli equation or can I simply find an integrating factor?

  • $\begingroup$ Are you sure its denominator isn't $1-4((y+1)^3)x$? $\endgroup$ – Ali Ashja' Sep 17 at 6:20
  • $\begingroup$ I am positive that 4 is actually a 3. I'm looking at it right now. Out of curiosity, and to gain an understanding, how important is that? $\endgroup$ – Fringe_Agent13 Sep 17 at 6:23
  • $\begingroup$ @Fringe_Agent13 Note you have mismatched brackets in the denominator, i.e., in $[1-3((y+1)^3)x$. In particular, you have $3$ left brackets and just $2$ right ones. Please check this & make the appropriate correction(s). Thanks. $\endgroup$ – John Omielan Sep 17 at 6:26
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    $\begingroup$ @Fringe_Agent13 If it was $1−4((y+1)^3)x$, the equation become a complete differential and solve easily. $\endgroup$ – Ali Ashja' Sep 17 at 7:00
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    $\begingroup$ In your original formulation with the $x$ inside, you get $\frac{y'}{y+1}=(x(y+1)^3)'$ so that $\ln|y+1|+C=x(y+1)^3$ and $C=0$ for the initial condition. $\endgroup$ – Dr. Lutz Lehmann Sep 17 at 7:12

The ODE you're asking to solve is

$$\frac{dy}{dx} = \frac{(y+1)^4}{[1-3((y+1)^3)]x}, \; y(0)=0 \tag{1}\label{eq1}$$

This is a separable ODE. To solve it, multiply both sides of \eqref{eq1} by $dx\left(\frac{1-3(y+1)^3}{(y+1)^4}\right)$ to get

$$\left(\frac{1-3(y+1)^3}{(y+1)^4}\right)dy = \frac{dx}{x} \tag{2}\label{eq2}$$

Integrating each side gives

$$\frac{1}{-3(y+1)^3} - 3\ln|y+1| = \ln|x| + C \tag{3}\label{eq3}$$

You can't use the initial condition to determine $C$ as $\ln(0)$ is not defined. Note that, on the RHS, for $y = 0$, you get $\frac{1}{-3}$. Also, in the original ODE in \eqref{eq1}, the initial condition has the numerator being $1$ but the denominator being $0$, so the derivative is not defined. As such, it appears the ODE is ill-posed.

  • $\begingroup$ Maybe I'm overlooking this but where did you use an integrating factor? I absolutely cannot find it. $\endgroup$ – Fringe_Agent13 Sep 17 at 6:46
  • $\begingroup$ @Fringe_Agent13 I updated my answer to explain it's a separable ODE. I separated the $x$ and $y$ parts and integrated them separately. There's no integrating factor as such, unless you want to consider what I multiply both sides by to be an integrating factor. $\endgroup$ – John Omielan Sep 17 at 6:47
  • $\begingroup$ I'm pretty sure this is Bernoulli. I can't use separation of variables. $\endgroup$ – Fringe_Agent13 Sep 17 at 7:03
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    $\begingroup$ @Fringe_Agent13 In that case, it seems you have not stated the equation correctly. If you had the denominator being $1 - 3(y+1)^3 x$ instead, then the initial condition would make sense. Alternatively, Ali Ashja''s comment suggests the denominator should be $1 - 4(y+1)^3 x$ instead. Regardless, I don't know about the equation being Bernoulli and using an integrating factor, but at least the ODE in either case wouldn't be ill-posed, as it is stated now. $\endgroup$ – John Omielan Sep 17 at 7:07
  • $\begingroup$ I am literally sitting here and looking at this paper. It is a 3. I think the way you have the denominator written is correct. I'm not sure. If you could do it with an integrating factor then I'm just going to go with that. I literally cannot use separation of variables because the professor said use integrating factor. Anything other method will cause me to lose points. $\endgroup$ – Fringe_Agent13 Sep 17 at 7:11

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