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I’m new to Laplace transforms and I’m trying to solve this question:

$$\dfrac{dy}{dx}=\dfrac{2\sqrt y}{(3-2x)(1-x)^2}$$

y = 0 when x = 0

Any help would be very much appreciated as I’ve researched online but it hadn’t made it any clearer.

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  • $\begingroup$ Do you absolutely need to use Laplace transforms? It seems much easier without it (the ODE is separable). $\endgroup$ – projectilemotion Sep 17 '19 at 10:57
  • $\begingroup$ @projectilemotion it was given as an extension question to use Laplace transforms but I’ve spent so long trying to learn how to apply it that I’m kind of giving up! $\endgroup$ – Valkiyare Sep 17 '19 at 10:59
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The Laplace transform can be applied only to linear DE's . The present DE is nonlinear but it can be linearized with the transformation

$$ y(x) = z^2(x) $$

because then

$$ y' = 2 z z' = \frac{2 z}{(3-2x)(1-x)^2} $$

or

$$ z' = \frac{1}{(3-2x)(1-x)^2} $$

which now is linear

now applying the Laplace transform to a non usual function (asking Wolfram)

$$ s Z(s) - z_0 = -\frac{1}{10} e^s \text{Ei}(-s)+\frac{2}{5} e^{-3 s/2} \Gamma \left(0,-\frac{3 s}{2}\right)-\frac{1}{2} e^{-s} \Gamma (0,-s) $$

and then after antitransforming

$$ z(t) = \frac{1}{10} \left(-4 \theta \left(t-\frac{3}{2}\right) \theta (-t) \log \left(\frac{3}{3-2 t}\right)-5 \theta (t-1) \theta (-t) \log (1-t)+\log (t+1)+10 z_0\right) $$

Here $\theta(\cdot)$ is the Heaviside theta function.

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(I'm sorry, but I don't have enough reputation to leave this a a comment, so I'm leaving a more detailed "answer")

Does the question ask to solve the ODE explicitly by using the Laplace transform? It'd be quite surprising if it does and I don't see a trivial way to do so. We usually solve ODEs by applying the Laplace transform for linear differential equations (not necessarily homogenous!), most popularly of order 2 or less. This isn't the case for the differential equation you want to solve- we don't have a linear relation between $y$ and it's derivative.

The most useful fact about the Laplace transform, the one that makes it useful to solve linear ODEs is the following relation between the Laplace transform and the derivative: $$\mathcal{L}(f')=s\mathcal{L}(f)-f(0)$$

Do note we also need and an initial value of $f$ evaluated at $0$. Another useful fact about the Laplace transform- since it's defined by the integral:

$$\mathcal{L}(f)=\int_{0}^\infty f(t)e^{-st}dt$$

It is a linear transformation! (This can be seen by using the linearity of the integral).

These properties of the Laplace transform actually reduce solving an IVP from a differential problem to an algebraic one (It is is a really rich field of research called Operational Calculus!). After applying the transform to both sides of the equation we re-arrange the equation to that we can use the last important property of the Laplace transformation: It is invertible! so after re-arranging we can apply $\mathcal{L}^{-1}$ to both sides and thus solve for $f$ .


If the question doesn't explicitly ask to solve the ODE by applying the Laplace transform , this ODE can actually be solved by separation of variables, and I'll give a little hint in case it wasn't obvious: $$\frac{dy}{dx}=\frac{2\sqrt{y}}{(3-2x)(1-x)^2}\Longrightarrow\frac{dy}{\sqrt{y}}=\frac{2dx}{(3-2x)(1-x)^2}$$

And then integrate on both sides, using partial fraction decomposition on the RHS.


If you really do have to use the Laplace transform It might help if you could provide us with the original problem so we'd have some context.

EDIT: I see you've updated the initial value, Thank you. I'll be trying to figure something out and if I come up with anything I'll make another edit.

EDIT 2: I'm adding a solution for reference. Please note I may have made some mistakes as well, so if you get something a bit different from me let me know in the comments and it might be best to check with other student for their results as well.

so this is the equation we get:

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

After partial fraction decomposition and some algebra we get:

$$2\sqrt{y}=\frac{-2}{x-1}-4\ln{|3-2x|}+4\ln{|x-1|}+C$$

Dividing both sides by 2 and taking the square of both sides to solve for y: $$y=\bigg(\frac{-1}{x-1}-2\ln{|3-2x|}+2\ln{|x-1|}+C\bigg)^2$$

Plugging the initial value to solve for C now:

$$0=(-1-2\ln{3}+C)^2\Longrightarrow C=1+2\ln{3}$$

and the solution to the IVP is: $$y=\bigg(\frac{-1}{x-1}-2\ln{|3-2x|}+2\ln{|x-1|}+1+2\ln{3}\bigg)^2$$

As I already said, I might have made some mistakes along the way, but this is the sketch for the solution. Good luck!

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  • $\begingroup$ Hello,the original question was a stated above: my teacher wrote it on the board and asked us to solve it using Laplace transforms as an extension. Additionally, y = 0 when x = 0. Your answer is great, thank you very much. If I can’t use LT I’ll resort to separating the variables. $\endgroup$ – Valkiyare Sep 17 '19 at 11:48
  • $\begingroup$ for the integration on the RHS, what do I do with the dx? Do I just discard it and integrate the rest of the RHS? (I'm quite new to Differential Equations and Laplace Transforms generally) $\endgroup$ – Valkiyare Sep 17 '19 at 14:36
  • $\begingroup$ You don't "discard" it but rather this dx is exactly what "let's" us integrate both sides of the equation. the dy means you integrate the LHS with respect to y and the dx means the RHS is integrated with respect to x. This is actually the mechanism behind separation of variables! (and don't forget these are indefinite integrals so you'll need +c's, you'll get rid of them after plugging the initial value once you're done.) $\endgroup$ – omer Sep 17 '19 at 14:46

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