Question

Differential Equations, Laplace Transform of derivatives and integrals. ODEs Question and answer check

Answer 1

Original Equation given by the professor for the respected section of Laplace transform:\[Y = \frac{ sy(0) +ay(0) +y'(0) +R}{ s^2+as+b }\]Equation of problem\[y'' -y'-6y=0\]wiith\[y(0)=11,y'(0)=28\]
Work\[Y = \frac{ 11s-11+28 }{ s^2-s-6 }\]\[=\frac{ 11(s-1)+28 }{ (s ^{2}-s+\frac{ 1 }{ 4 }) -6 -\frac{ 1 }{ 4 }}\]\[=\frac{ 11(s-1) +28 }{ (s-\frac{ 1 }{ 2 })^{2}-\frac{ 25 }{ 4 } }\]\[=\frac{ 11(s-\frac{ 1 }{ 2 })+28-\frac{ 11 }{ 2 }}{ (s-\frac{ 1 }{ 2 })^{2}-\frac{ 25 }{ 4 } }\]\[=\frac{ 11(s-\frac{ 1 }{ 2 }) +\frac{ 45 }{ 2 }}{ (s-\frac{ 1 }{ 2 })^{2}-\frac{ 25 }{ 4 } }\]\[=\frac{ 11(s-\frac{ 1 }{ 2 }) }{ (s-\frac{ 1 }{ 2 })^{2} -\frac{ 25 }{ 4}}+\frac{ \frac{ 45 }{2} }{ (s-\frac{ 1 }{ 2}) ^{2}-\frac{ 25 }{ 4}}\]here I establish the s-shift F(s-a) , where a = 1/2 and then the Answer is:\[y = e ^{0.5t}(11\cosh2.5t+9\sinh2.5t)\]can someone check if I'm right?

Answer 2

the 2 last lines somehow are weird. to me, the first term ok, the second term seems odd . numerator is 45/2 when b is denominator is 25/4
let the first term a side, I try on second term by back to the original form of denominator, I get \[\frac{\frac{45}{2}}{(s-3)(s+2)}\]
by partial fraction, I get it is \[ \frac{9}{2}\frac{1}{s-3} -\frac{9}{2}\frac{1}{s+2}\]
take inverse Laplace for it , I got \[\frac{9}{2}e^3t -\frac{9}{2}e^{-2t}\]
combine with your first term, the whole solution is \[11e^{0.5t}cosh 2.5t +\frac{9}{2}e^3t -\frac{9}{2}e^{-2t}\]

Answer 3

Hi Loser66, thank you for responding, the way I did my 2nd term is that the denominator is 25/4 which resorted to (5/2)^2 as a square so it gave me a laplace transform of (sinh at), with the constant (45/2) on the outside, of a/(s^2-a^2)

\[\frac{ \frac{ 45 }{ 2} }{ \frac{ 5 }{ 2 }}\frac{ \frac{ 5 }{ 2 } }{ (s-\frac{ 1 }{ 2 })^{2}-(\frac{ 5 }{ 2})^{2}}\] and that'show I got my answer. I wonder if that seems to clear it up for you on how I did it

Answer 4

ok