Solving Shifted Data IVP with Laplace transform

Solution 1:

Well, using Laplace transform it is not hard to see that:

$$\mathscr{L}_x\left[\text{y}''\left(x\right)\right]_{\left(\text{s}\right)}=\text{s}^2\cdot\text{Y}\left(\text{s}\right)-\text{s}\cdot\text{y}\left(0\right)-\text{y}' \left(0\right)\tag1$$

And:

$$\mathscr{L}_x\left[2x\right]_{\left(\text{s}\right)}=\frac{2}{\text{s}^2}\tag2$$

So, we get:

$$\text{s}^2\cdot\text{Y}\left(\text{s}\right)-\text{s}\cdot\text{y}\left(0\right)-\text{y}' \left(0\right)=\frac{2}{\text{s}^2}\space\Longleftrightarrow\space\text{Y}\left(\text{s}\right)=\frac{2}{\text{s}^4}+\frac{\text{y}\left(0\right)}{\text{s}}+\frac{\text{y}' \left(0\right)}{\text{s}^2}\tag3$$

Using inverse Laplace transform, we can see that:

$$\text{y}\left(x\right)=\frac{x^3}{3}+\text{y}\left(0\right)+\text{y}' \left(0\right)\cdot x\tag4$$

using your initial conditions, we get:

$$ \begin{cases} \text{y}\left(\frac{\pi}{4}\right)=\frac{1}{3}\cdot\left(\frac{\pi}{4}\right)^3+\text{y}\left(0\right)+\text{y}' \left(0\right)\cdot\frac{\pi}{4}=\frac{\pi}{2}\\ \\ \text{y}'\left(\frac{\pi}{4}\right)=\left(\frac{\pi}{4}\right)^2+\text{y}' \left(0\right)=2-\sqrt{2} \end{cases}\tag5 $$

So, this gives:

$$ \begin{cases} \text{y}\left(0\right)=\frac{\pi}{2}-\frac{1}{3}\cdot\left(\frac{\pi}{4}\right)^3-\left(2-\sqrt{2}-\left(\frac{\pi}{4}\right)^2\right)\cdot\frac{\pi}{4}=\frac{\pi\left(24\sqrt{2}+\pi^2\right)}{96}\\ \\ \text{y}' \left(0\right)=2-\sqrt{2}-\left(\frac{\pi}{4}\right)^2 \end{cases}\tag6 $$

So, the solution is:

$$\text{y}\left(x\right)=\frac{x^3}{3}+\frac{\pi\left(24\sqrt{2}+\pi^2\right)}{96}+\left(2-\sqrt{2}-\left(\frac{\pi}{4}\right)^2\right)x\tag7$$