# Is there a Laplace transform for an unknown squared function?

I was solving the differential equation $$y'-y = e^x y^2$$ I know that this is a simple differential equation, and I would solve it with a simple change of the variable, however, I was wondering if I could use Laplace transform to make it easier and more direct. I asked my professor, and she said she didn't know.

So is there a way to do $$\mathscr{L}([y(x)]^2)$$?

• No. The main property of the Laplace transform is that it is a linear transformation. This tells you how it acts for sums and for scalar multiples. But not for products, not even for squares. Mar 18, 2021 at 21:52

## 3 Answers

$$y'-y = e^x y^2$$ Substitute $$u=-\dfrac 1y$$ then solve with Laplace transform: $$u'+u=e^x$$ Otherwise you can't use Laplace transform with non linear DE.

Suppose $$y$$ is analytic at $$0$$, having power series $$y(x) = \sum_{k=0}^\infty a_k x^k \text{.}$$ Then $$y^2(x) = \sum_{n=0}^\infty \left( \sum_{i=0}^n a_i a_{n-i} \right) x^n$$ and $$\mathcal{L}\{y^2\}(s) = \sum_{n=0}^\infty \left( \sum_{i=0}^n a_i a_{n-i} \right) \frac{n!}{s^{n+1}} \text{,}$$ wherever that series converges (if it converges anywhere).

I don't imagine this would ever be useful. (Just how badly does applying the Frobenius method have to go to try taking a Laplace transform in the middle of it?) But, a calculation can be done and there is a (formal) result.

One can certainly expand somewhere other that $$0$$, but this doesn't improve the situation.

• I guess there is some kind of convolution hidden in the final formula. I agree that it is unlikely to be useful but it is a nice exploration. Mar 18, 2021 at 23:03
• @GiuseppeNegro : Yes, multiplication of power series yields convolution of the coefficients. Mar 18, 2021 at 23:10

Unless if you are able to make a clever substitution you cannot solve this ODE with the Laplace Transform.

The Laplace Transform is only used for solving linear equations, as it's a linear transform.

Trust me, i have tried solving nonlinear equations, i have failed.

What a Laplace Transform does to an ODE is that it basically finds the characteristic polynomial for a linear ODE.

$$ay''(t)+by'(t)+cy(t)=f(t)$$

# $$\downarrow\mathscr{L}$$

$$a[s^2Y(s)-sy(0)-y'(0)]+b[sY(s)-y(0)]+cY(s)=F(s)$$

where $$y(0), y'(0)$$

are the intial conditions, as we use the Laplace Transform to see how a system dampens, so we always need those.

As you can see we get an equation of the form:

$$Y(s)=\xi(s)$$

where $$\xi$$ is a polynomial made of $$s$$-s.

For $$y(0)=0, y'(0)=0$$

The above equation will result in:

$$Y(s)=\dfrac{F(s)}{as^2+bs+c}$$

# $$\downarrow\mathscr{L}^{-1}$$

Now after transforming back we'll get the solution to $$y(t)$$ and of course we get the same solution as if we use the classical method to solve the ODE.

Edit: For your case even with a substitution you'll still have to plug in initial conditions which sadly make this method kind of bad for general solutions (or you could assume y(0)=y'(0)=0, but that will give you a solution to an IVP)