How do I apply a Laplace transform to solve this ODE?

Question

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.

Answer 1

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.

Answer 2

(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 $f$ .

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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.

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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!