Solving Differential Equations Using Laplace Transform in Five Steps

The long equation terms of Laplace transform (LT) or inverse LT example and problems looks scary and wild beast in the jungles of mathematics, right? But believe me you can tame it actually to kill even scarier beast called differential equations of higher orders.  And the higher order DEs are extremely useful for creating mathematical model of engineering design of systems.

Before explaining the steps for solving a differential equation example, see how the overall procedure works:

  • The differential equation (with initial value points or IVP) are transformed to algebraic equations using the laplace transform because of the fact that finding solution is much easier for algebraic equations than differential equations.
  • The solution of the transformed algebraic equation is found.
  • This solution is then inverse transformed to get the solution of the actual differential equation.

You will need the following two formulas:

L (y)= F(s)= ∫0 e-st y dt………………………………………………………………eqn.1

Where,

L(y) is the Laplace transform (LT) of y

y is a function of t

s is the new variable of the transform

 

 

L(y(n)) = sn L(y) – sn-1 y(0) – sn-2 y(1)(0) – s y(n-2)(0) – y(n-1)(0) ………….eqn.2

Where,

y(n) is the nth order derivative of the function y

L(y(n)) is the LT of y(n)

y is a function of t

s is the new variable of the transform

 

 

L{ay1 + by2} = aL(y1) + bL(y2)…………………………………….eqn.3

Where,

y1, y2 are function of some common variable (say t)

L(y1), L(y2) are the LT of y1 and y2 respectively.

 

 

Example: we have to find the solution for

y(2)+3 y(1)+2y=et

y(0)=1

y(1)(0)=-1

 

Solution:

Step-1: By applying the eqn.3 we will get

L(y(2))+3 L(y(1))+2L(y)=L(et)…………………….s.1

 

Step-2: From the eqn.2 we can write that

 

L(y(2)) = s2L(y) – sy(0) – y(1)(0)

Since, y(0)=1 and y(1)(0)=-1 hence,

L(y(2))  = s2L(y) – s +1………………………………s.2

Also,

L(y(1))=sL(y) – y(0)

Since, y(0)=1 hence,

L(y(1))=sL(y) – 1…………………………………………s.3

 

By using the equation s.2 and s.3, the equation s.1 can be written as:

s2L(y) – s +1 + 3sL(y) – 3 + 2L(y) = L(et)

or,  s2L(y) + 3 sL(y) + 2L(y) –s -2 = L(et) …………………s.4

 

Step-3: L(t) can be found out by using the eqn.1 as below:

L(et) = 0 e-st et dt= 1/(s-1)………………………s.5

 

From the equation s.4 and s.5 it can be written as:

s2L(y) + 3s L(y) + 2L(y) –s -2 = 1/(s-1)

or, L(y)= (s2+s-2)/[(s-1)(s+2)(s+1)]……………..s.6

Step-4: Now we have to apply the partial decomposition technique, in order to put the equation s.6 in a form suitable for inverse LT.

Let’s say,

(s2+s+2)/[(s-1)(s+2)(s+1)] = A/(s-1) + B/(s+2) +C/(s+1)…..s.7

Or, (s2+s+2)=A(s+2)(s+1) + B(s-1)(s+1) + C(s+2)(s-1)

 

Now, by trying different arbitrary values of s, we get

For, s=1   A=4/6

For, s=-1  C=-1

For, s=-2  B=4/3

 

So, the equation s.6 can be rewritten as

L(y)=(4/6)/(s-1) + (4/3)/(s+2) + (-1)/(s+1)………..s.8

Step-5: Now we can find the solution in terms of t by performing inverse transform of the equation s.8. We will use the standard table for finding the inverse transform of the quantities of the equation s.8

From the standard table,

the inverse transform of

(4/6)/(s-1) is (4/6)et

(4/3)/(s+2) is (4/3)e-2

(-1)/(s+1) is (-1)e-1

So, the solution in terms of t is

Y(t)= (4/6)et + (4/3)e-2 – e-1

 

 

References:

Paul’s Online Math Notes

SOS math

 

Related books:

 

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