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Find a solution to the following ODE using Laplace tranforms

$$xf''(x) + 2f'(x) + xf(x) = 0$$

Find a second independent solution and explain why it was not found using Laplace transforms.

Edit: My initial attempt:

Applying the transformation I got:

$$-\frac{d}{ds}[F(s)-sf(0)-f'(0)] + 2[sF(s)-f(0)] - \frac{d}{ds}[F(s)] = 0 $$

$$F'(s) =\frac{f(0)-1}{s^2+1}$$

$$F(s) = (f(0) - 1)tan^{-1}(s)$$ Where $F(s)$ is the Laplace transform of $f(s)$. My problem was then that I could not invert this. I'm not sure if I just can't find the inversion or my steps were wrong somewhere.

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$$xf''(x) + 2f'(x) + xf(x) = 0$$ NOTE:

The Laplace method uses $f(0)$. This implies that $f(0)$ must be finite. As a consequence, if an ODE has a solution $f(x)$ which is not finite at $x=0$, the Laplace method will fail to find this solution.

For example, suppose that the general solution of the ODE be $$f(x)=c_1\frac{\sin(x)}{x}+c_2\frac{\cos(x)}{x}$$ then, the Laplace method will failed to find this general solution. Only the solution with $c_2=0$ can be found because $\left(\frac{\cos(x)}{x}\right)_{x=0}=\pm\infty$.

The solution found, thanks to the Laplace transform method, is (see below): $$f(x)=f(0)\frac{\sin(x)}{x}$$ SOLVING:

enter image description here

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  • $\begingroup$ That's brilliant, thank you! $\endgroup$ – Fahrenheit997 Jan 28 '17 at 15:20
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Use $${\mathcal L}^{-1}\Big(\int_0^\infty F(s)ds\Big)=\frac{{\mathcal L}^{-1}(F(s))}{x}=\frac{f(x)}{x}$$ which $$\arctan s=\dfrac{\pi}{2}-\int_s^\infty\frac{1}{1+t^2}dt$$

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