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I was hoping someone can help me with the following.

I had to solve the following ODE and get the implicit form.

$$\frac{dy}{dx}=\frac{1+3y}{y} * x^2$$

this give by separating and integrating the following: $1/9*(3y-ln(3y+1))=1/3*x^3 +c $ which can also be written as (c is an arbitrary constant): $$3y-ln(3y+1)=3x^3+c $$

I don't know exactly how to solve this, I saw something about lambert functions but that wasn't entirely clear to me. My start was rewriting the equation as:

$3y+1-ln(3y+1)=3x^3+c$
$u-ln(u)=3x^3+c$
$ln(e^u)-ln(u)=3x^3+c$ or $ln(e^{-u})+ln(u)=-3x^3+c$
$ln(\frac{1}{u}*e^u)=3x^3+c$ or $ln(u*e^{-u})=-3x^3+c$
$\frac{1}{u}*e^u= e^{(3x^3+c)}$ or $u*e^{-u}=e^{-3x^3+c}$
but that's as far as I can get. Since I don't know how to apply the lambert function for that nor any other way to solve it.

Any help with this would be greatly appreciated.

Bob van de Voort
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2 Answers2

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By definition of the Lambert W function: if $ze^z=A$ than $z=W(A)$.

so, if we have: $$ ue^{-u}=e^{-3x^2+c} $$ that is : $$ -ue^{-u}=-e^{-3x^2+c} $$ we have: $$ -u=W\left(-e^{-3x^2+c} \right)\quad \iff \quad u=-W\left(-e^{-3x^2+c} \right) $$

so, for $u=3y+1$ we find: $$ y=\frac{1}{3}\left( -W\left(-e^{c-3x^2} \right)-1\right) $$

Emilio Novati
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1

Well, we have that:

$$\text{y}'\left(x\right)=\frac{1+\text{n}\cdot\text{y}\left(x\right)}{\text{y}\left(x\right)}\cdot x^2\space\Longleftrightarrow\space\int\frac{\text{y}\left(x\right)\cdot \text{y}'\left(x\right)}{1+\text{n}\cdot\text{y}\left(x\right)}\space\text{d}x=\int x^2\space\text{d}x\tag1$$

For a constant $\text{n}$

Now, we use:

  • Substitute $\text{u}=\text{y}\left(x\right)$: $$\int\frac{\text{y}\left(x\right)\cdot \text{y}'\left(x\right)}{1+\text{n}\cdot\text{y}\left(x\right)}\space\text{d}x=\int\frac{\text{u}}{1+\text{n}\cdot\text{u}}\space\text{d}\text{u}=\frac{\text{n}\cdot\text{u}-\ln\left|1+\text{n}\cdot\text{u}\right|}{\text{n}^2}+\text{C}_1\tag2$$
  • $$\int x^2\space\text{d}x=\frac{x^3}{3}+\text{C}_2\tag3$$

So, we get:

$$\frac{\text{n}\cdot\text{y}\left(x\right)-\ln\left|1+\text{n}\cdot\text{y}\left(x\right)\right|}{\text{n}^2}=\frac{x^3}{3}+\text{C}\tag4$$

Simplifying it a bit:

$$\text{n}\cdot\text{y}\left(x\right)-\ln\left|1+\text{n}\cdot\text{y}\left(x\right)\right|=\text{n}^2\cdot\frac{x^3}{3}+\text{C}\tag5$$

Assuming that:

$$\left(1+\text{n}\cdot\text{y}\left(x\right)\right)\in\mathbb{R}^+\tag6$$

We can write:

$$\text{n}\cdot\text{y}\left(x\right)-\ln\left(1+\text{n}\cdot\text{y}\left(x\right)\right)=\text{n}^2\cdot\frac{x^3}{3}+\text{C}\tag7$$

And now we can use a general property:

$$\text{a}\cdot\text{p}-\ln\left(\text{b}+\text{c}\cdot\text{p}\right)=\text{d}\space\Longleftrightarrow\space\text{p}=-\frac{\text{a}\cdot\text{b}-\text{c}\cdot\mathcal{W}\left(-\frac{\text{a}\cdot\exp\left(-\frac{\text{a}\cdot\text{b}}{\text{c}}-\text{d}\right)}{\text{c}}\right)}{\text{a}\cdot\text{c}}\tag8$$

Jan Eerland
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