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How to solve second-order ODEs using Laplace Transform method, Partial Fraction Decomposition
You can use the Laplace transform method to solve second-order ODEs.
This method can be applied so that you don’t need to stop and figure out which method to use; however, Laplace Transform is usually done in the second-half of a introductory differential equations course.
You do need a cheat sheet of tables to use Laplace Transforms, so ask your instructor how they feel about this method.
You may be required to use other methods, such as integrating factors and homogeneous and particular solutions for the earlier parts of your course.
Note, you need the initial conditions to be able to use the Laplace transform, this will become apparent quickly!
Example problem: Solve \(y” + y = 3\sin \left( {2t} \right)\) using Laplace transform, also given initial conditions \(y(0) = 6\) and \(y'(0) = 1\).
Define the Laplace transform operator, use capital letters to denote this.
Let \(L\left[ {y(t)} \right] = Y(s)\)
Perform the Laplace operation on each term, it is a linear operator.
\[L\left[ {y”} \right] + L\left[ y \right] = 3L\left[ {\sin \left( {2t} \right)} \right]\]
Now use a Laplace Table to get these values quickly! You will need to be careful with multiplying, and the frequency of the sin term here is “2” so be careful.
\[{s^2}Y – sy(0) – y'(0) + Y = 3\left( {\frac{2}{{{s^2} + 4}}} \right)\]
Substitute in the initial conditions!
\[{s^2}Y – 6s – 1 + Y = \left( {\frac{6}{{{s^2} + 4}}} \right)\]
The trick is to isolate for Y.
\[Y(1 + {s^2}) = \frac{6}{{{s^2} + 4}} + 1 + 6s\]
Now divide by \((1 + {s^2})\)
\[Y = \frac{{6s}}{{(1 + {s^2})}} + \frac{1}{{(1 + {s^2})}} + \frac{6}{{\left( {{s^2} + 4} \right)(1 + {s^2})}}\]
You will need to use the inverse Laplace operation to find the solution to the ODE, but you can’t find the inverse Laplace of a fraction with two factors in the denominator.
So use “partial fraction decomposition” to split that fraction up, you need constants for up to linear terms in both numerators.
\[\frac{6}{{\left( {{s^2} + 4} \right)(1 + {s^2})}} = \frac{{As + B}}{{\left( {{s^2} + 4} \right)}} + \frac{{Cs + D}}{{\left( {{s^2} + 1} \right)}}\]
Multiply by the denominators, \(\left( {1 + {s^2}} \right)\left( {{s^2} + 4} \right)\) to get:
\[6 = (As + B)({s^2} + 1) + (Cs + D)({s^2} + 4)\]
\[6 = A{s^3} + As + B{s^2} + B + C{s^3} + 4Cs + D{s^2} + 4D\]
To generate four equations easily, equate both sides by each power.
For s terms to the third power: 0 = A + C
For s terms to the second power: 0 = B + D
For s terms to the first power: 0 = A + 4C
For s terms to the 0th power: 6 = B + 4D
It’s not really complicated, B = -D, then use above equation to get D = 2, B = -2, and A = -C, 0 = 3C, C = 0, A = 0, just keep manipulating to get the values. Then plug them into the Y equation from previously. Then,
\[Y = \frac{{6s}}{{(1 + {s^2})}} + \frac{1}{{(1 + {s^2})}} + \frac{2}{{{s^2} + 1}} – \frac{2}{{{s^2} + 4}}\]
Now take the inverse Laplace of each fraction, use a Laplace table if you want.
For example, consider the last fraction, factor out terms,
\[ – 2{L^{ – 1}}\left( {\frac{1}{{{s^2} + {2^2}}}} \right)\]
And the inverse Laplace of that would result in \( – \sin (2t)\)
And just find the inverse Laplace of all the terms, to get the solution:
\[y(t) = 3\sin (t) + 6\cos (t) – \sin (2t)\]