Solve Differential Equations Using Laplace Transform
Examples of how to use Laplace transform to solve ordinary differential equations (ODE) are presented. One of the main advantages in using Laplace transform to solve differential equations is that the Laplace transform converts a differential equation into an algebraic equation.
Heavy calculations involving decomposition into partial fractions are presented in the appendix at the bottom of the page.
Example 1
Use Laplace transform to solve the differential equation
−2y′+y=0
with the initial conditions y(0)=1 and y is a function of time t.
Solution to Example1
Let Y(s) be the Laplace transform of y(t)
Take the Laplace transform of both sides of the given differential equation: L{y(t)}=Y(s)
L{−2y′+y}=L{0}
Use linearity property of Laplace transform to rewrite the equation as
−2L{y′}+L{y}=L{0}
Use derivative property to rewrite the term L{y′}=(sY(s)−y(0)).
−2(sY(s)−y(0))+Y(s)=0
Expand the above as
−2sY(s)+2y(0)+Y(s)=0
Substitute y(0) by its given numerical value
−2sY(s)+2+Y(s)=0
Solve the above for Y(s)
Y(s)(1−2s)=−2
Y(s)=22s−1
Y(s)=1s−1/2
We now use formula
(3) in the table of formulas of Laplace transform to find the inverse Laplace transform of Y(s) obtained above as
y(t)=e12t
Note: Check solution
let's check that the solution obtained y(t)=e12t satisfies the given differential equation
−2y′+y=−2((1/2)e12t)+e12t
Simplify the above
−e12t+e12t=0 ; differential equation satisfied.
y(0)=e120=e0=1 ; initial value also satisfied.
Example 2
Use Laplace transform to solve the differential equation
y′′−2y′−3y=0
with the initial conditions y(0)=2 and y′(0)=−1 and y is a function of time t.
Solution to Example 2
Let Y(s) be the Laplace transform of y(t)
Take the Laplace transform of both sides of the given differential equation
L{y′′−2y′−3y}=L{0}
Use linearity property of Laplace transform to rewrite the equation as
L{y"}−2L{y′}−3L{y}=L{0}
Use first and second derivative properties to rewrite the terms L{y"} and L{y′} and simplify the right side.
s2Y(s)−sy(0)−y′(0)−2(sY(s)−y(0))−3Y(s)=0
Substitute y(0) and y′(0) by their numerical values and expand
numerical values and expand
s2Y(s)−2s+1−2sY(s)+4−3Y(s)=0
Group like terms and keep terms with Y(s) on the left
s2Y(s)−2sY(s)−3Y(s)=2s−5
Factor Y(s) out
Y(s)(s2−2s−3)=2s−5
Solve the above for Y(s)
Y(s)=2s−5s2−2s−3
Expand the right side into partial fractions (see details in Appendix A at the bottom of the page)
Y(s)=74(s+1)+14(s−3)
We now use formula (3) in the table of formulas of Laplace transform to find the inverse Laplace transform of Y(s) which is given by
y(t)=74e−t+14e3t
You may check that the solution obtained satisfies the differential equation and the initial values given.
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