Laplace Transforms: y'' - 6y' +13y = 0 ; y(0)=0, y' (0)= -3
- Thread starter r1luher0
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Re: Laplace Transforms
I will step through this one so you can do the others on your own. OK?
LaPlace transforms turn a DE into an algebraic, so we can solve easier.
\(\displaystyle y'=pY-y(0)\)
\(\displaystyle y''=p^{2}Y-py(0)-y'(0)\)
Sub these into your DE:
\(\displaystyle p^{2}Y-py(0)-y'(0)-6(pY-y(0))+13Y=0\)
Sub in the initial conditions given, factor out Y, and get:
\(\displaystyle Y(p^{2}-6p+13)=-3\)
\(\displaystyle Y=\frac{-3}{p^{2}-6p+13}\)
Now, look this up in a table to LaPlace transforms and we get:
\(\displaystyle \boxed{y=\frac{-3}{2}e^{3t}sin(2t)}\)
For the last one, find the laplace of t^2. Which is \(\displaystyle \frac{2}{p^{3}}\)
\(\displaystyle pY-y(0)+5Y=\frac{2}{p^{3}}\)
\(\displaystyle pY+5Y=\frac{2}{p^{3}}\)
\(\displaystyle Y=\frac{2}{p^{3}(p+5)}\)
Taking partial fractions:
\(\displaystyle Y=\frac{-2}{125(p+5)}+\frac{2}{125p}-\frac{2}{25p^{2}}+\frac{2}{5p^{3}}\)
Now, these are relatively easier ones by looking them up in the Laplace table.
I will step through this one so you can do the others on your own. OK?
LaPlace transforms turn a DE into an algebraic, so we can solve easier.
\(\displaystyle y'=pY-y(0)\)
\(\displaystyle y''=p^{2}Y-py(0)-y'(0)\)
Sub these into your DE:
\(\displaystyle p^{2}Y-py(0)-y'(0)-6(pY-y(0))+13Y=0\)
Sub in the initial conditions given, factor out Y, and get:
\(\displaystyle Y(p^{2}-6p+13)=-3\)
\(\displaystyle Y=\frac{-3}{p^{2}-6p+13}\)
Now, look this up in a table to LaPlace transforms and we get:
\(\displaystyle \boxed{y=\frac{-3}{2}e^{3t}sin(2t)}\)
For the last one, find the laplace of t^2. Which is \(\displaystyle \frac{2}{p^{3}}\)
\(\displaystyle pY-y(0)+5Y=\frac{2}{p^{3}}\)
\(\displaystyle pY+5Y=\frac{2}{p^{3}}\)
\(\displaystyle Y=\frac{2}{p^{3}(p+5)}\)
Taking partial fractions:
\(\displaystyle Y=\frac{-2}{125(p+5)}+\frac{2}{125p}-\frac{2}{25p^{2}}+\frac{2}{5p^{3}}\)
Now, these are relatively easier ones by looking them up in the Laplace table.
mmm4444bot
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If you cannot find a Laplace table including Y(p) = -3/(p^2 - 6p + 13), then complete the square in the denominator first, and use the following basic Laplace transform inverse.
The inverse of Y(p) = b/[(p - a)^2 + b^2] is f(t) = e^(at) * sin(bt)
Completing the square on p^2 - 6p + 13 gives (p - 3)^2 + 4.
Y(p) = -3/[(p - 3)^2 + 4]
In order to match the form of the symbolic Y(p) with parameters a and b (shown above), we need the numerator to be 2, instead of -3. We accomplish this change by factoring out -3/2.
(-3/2) * 2/[(p - 3)^2 + 4] = -3/[(p - 3)^2 + 4]
f(t) = (-3/2) * e^(3t) * sin(2t)
The inverse of Y(p) = b/[(p - a)^2 + b^2] is f(t) = e^(at) * sin(bt)
Completing the square on p^2 - 6p + 13 gives (p - 3)^2 + 4.
Y(p) = -3/[(p - 3)^2 + 4]
In order to match the form of the symbolic Y(p) with parameters a and b (shown above), we need the numerator to be 2, instead of -3. We accomplish this change by factoring out -3/2.
(-3/2) * 2/[(p - 3)^2 + 4] = -3/[(p - 3)^2 + 4]
f(t) = (-3/2) * e^(3t) * sin(2t)
D
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To see the true advantage of Laplace's Method - solve this equation by regular method and compare the effort you had to put in find the homogeneous solution (no particular solution) and the apply the initial conditions to find the constants.