(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]t y'' + (3 t-1) y' + 3 y = 6 e^{-3t}[/tex]

y(0) = 1

y(5) = 2

2. Relevant equations

3. The attempt at a solution

I tried applying the Laplace transform to the equation but I was having a little trouble...

L [t·y''] = -dL[y'']/ds = s^{2}·Y' + 2·s·Y - y(0) = s^{2}·Y' + 2·s·Y - 1

L [t·y'] = -dL[y']/ds = s·Y' + Y

Therefore:

[tex]s^2 Y' + 2 s Y - 1 + 3 s Y' +3 Y - s Y + 1+ 3 Y = \frac{6}{s+3}[/tex]

[tex](s^2+3s)Y'+(s+6)Y=\frac{6}{s+3}[/tex]

And then solve the Y(s) and then use the inverse transform.

But it doesn't yield the correct result. I did the rest on the computer so I guess it's when I apply the transform where I make a mistake.

Thanks for your help.

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# Differential equation with Laplace transform

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