Key to "Solving Shifted Data IVP with Laplace transform{y″+y=2ty(π4)=π2y′(π4)=2−(2)"

Octavio Chen

Answered question

2022-03-31

Solving Shifted Data IVP with Laplace transform

{\begin{cases} y^{″} + y = 2 t \\ y (\frac{π}{4}) = \frac{π}{2} \\ y^{'} (\frac{π}{4}) = 2 - \sqrt{(} 2) \end{cases}

Answer & Explanation

undodaonePvopxl24

Beginner2022-04-01Added 13 answers

Step 1
Well, using Laplace transform it is not hard to see that:
1) $L_{x} {[y^{''} (x)]}_{(s)} = s^{2} \cdot Y (s) - s \cdot y (0) - y^{'} (0)$
And
2) $m a t h s c r {L}_{x} {[2 x]}_{(s)} = \frac{2}{s^{2}}$
So, we get:
3) $s^{2} \cdot Y (s) - s \cdot y (0) - y^{'} (0) = \frac{2}{s^{2}} \leftrightarrow Y (s) = \frac{2}{s^{4}} + \frac{y (0)}{s} + \frac{y^{'} (0)}{s^{2}}$
Using inverse Laplace transform, we can see that:
4) $y (x) = \frac{x^{3}}{3} + y (0) + y^{'} (0) \cdot x$
using your initial conditions, we get:
5)
${\begin{cases} y (\frac{π}{4}) = \frac{1}{3} \cdot (\frac{π}{4})^{3} + y (0) + y^{'} (0) \cdot \frac{π}{4} = \frac{π}{2} \\ y^{'} (\frac{π}{4}) = (\frac{π}{4})^{2} + y^{'} (0) = 2 - \sqrt{2} \end{cases}$
So, this gives:
6)
${\begin{cases} y (0) = \frac{π}{2} - \frac{1}{3} \cdot (\frac{π}{4})^{3} - (2 - \sqrt{2} - (\frac{π}{4})^{2}) \cdot \frac{π}{4} = \frac{π (24 \sqrt{2} + π^{2})}{96} \\ y^{'} (0) = 2 - \sqrt{2} - (\frac{π}{4})^{2} \end{cases}$
So, the solution is:
7) $y (x) = \frac{x^{3}}{3} + \frac{π (24 \sqrt{2} + π^{2})}{96} + (2 - \sqrt{2} - {(\frac{π}{4})}^{2}) x$

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