How to settle initial value difficulties making use of Laplace transforms

To use a Lalocation transform to deal with a second-order nonhomogeneous differential equations initial worth problem, we’ll have to use a table of Lalocation transcreates or the definition of the Laplace transcreate to put the differential equation in terms of ???Y(s)???.

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Once we settle the resulting equation for ???Y(s)???, we’ll desire to simplify it until we recognize that the terms in our equation match formulas in a table of Lalocation transcreates. Then we’ll make reverse substitutions for ???s??? in regards to ???t???.


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Given a differential equation and also initial problems, usage a table of Lalocation transcreates or the meaning to fix the initial value problem


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Step-by-step example for resolving the initial value trouble via a table of Laarea transforms


Use a Laplace transform to deal with the differential equation.


via ???y(0)=-1??? and also ???y"(0)=2???

To solve this difficulty utilizing Laarea transforms, we will should transform eincredibly term in our offered differential equation. From a table of Lalocation transforms, we have the right to redefine each term in the differential equation.





Plugging the transformed values ago right into the original equation gives


Now we’ll plug in the given initial conditions ???y(0)=-1??? and ???y"(0)=2???.





From below we want to solve for ???Y(s)??? so that we can use a reverse Laarea transcreate to readjust this equation into an equation for ???y(t)???.



???Y(s)left(s^2-10s+9 ight)=frac5+12s^2-s^3s^2???





We’ll must usage a partial fractions decomplace.




???5+12s^2-s^3=Asleft(s^2-10s+9 ight)+Bleft(s^2-10s+9 ight)???

???+Cleft(s^3-s^2 ight)+Dleft(s^3-9s^2 ight)???



???5+12s^2-s^3=left(As^3+Cs^3+Ds^3 ight)+left(-10As^2+Bs^2-Cs^2-9Ds^2 ight)???

???+left(9As-10Bs ight)+9B???

???5+12s^2-s^3=left(A+C+D ight)s^3+left(-10A+B-C-9D ight)s^2???

???+left(9A-10B ight)s+9B???

Equating coefficients, we gain a device of straight equations.

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Solving the fourth equation for ???B??? gives


Plugging this into the third equation gives

???9A-10left(frac59 ight)=0???





Plugging the values we’ve uncovered for ???A??? and also ???B??? into the first two equation gives


???-10left(frac5081 ight)+frac59-C-9D=12???

which is



which is



which is



which is



Adding<1>and<2>together gives






Plugging<3>into<1>we get





Plugging the values we discovered for ???A???, ???B???, ???C??? and also ???D??? earlier into the partial fractions decomposition will offer us


We’ll rearrange each term in the decomplace to make it easier to find a equivalent formula in the Laplace transcreate table.

???Y(s)=frac5081left(frac1s ight)+frac59left(frac1s^2 ight)+frac3181left(frac1s-9 ight)-2left(frac1s-1 ight)???

The terms staying inside the parentheses need to remind us of these transformations: