Diff. Eq. - Solve the following equation using Laplace transforms

Exmpl. #1) Solve the following equation using Laplace transforms. Where y(0) = 0, y'(0) = 0 and x(t) is a step change of magnitude 2.

Standard procedure of these problems, taking the Laplace transform of both sides!

But let's start with the left side first:

(*Laplace transforms of derivatives must be known before calculating... click here!)

After plugging in f(0) and f'(0), and then simplifying, we get:

Now let's do the Laplace transform of the right side of the equation. Remember that it was given to us that x(t) is a step change of magnitude 2. Laplace transform of a step function is always 1/s, and we will multiply it by 2 to represent the change of magnitude.

Time to factor out Y, which represents the Laplace transform of a function y(t).

---> (*Apply a/b/c/1 fraction rule!)

Now, we will use partial fraction decomposition to solve for Y.

Let's set this up.

Now distribute and simplify.

Group and equate the terms by power (right and left side of the equal sign)

You will find that,

A = 1/3

B = -1

C =
2/3

Now we take the inverse laplace transform of the equation to get our answer!

A standard laplace transform table, should help you easily calculate these values

*Inv. Laplace of variable is the same variable, but in the time domain. So, Inverse Laplace of Y is y(t)

and here is our answer:

Diff. Eq. - The Laplace Transform - Exmpl. 6

 Example 6 , The Laplace transform of the function f(t) = e^(at)

Diff. Eq. - The Laplace Transform

Chapter 8 - Laplace transforms

Laplace transforms are a tool for solving initial value problems where the differential equation is linear and has constant coefficients. 

The idea is to transform the differential equation into an algebraic equation whose solution is easier to obtain and then apply the inverse transform to solve the differential equation.

f(t)  →  F(s)

*f(t) being the original function

*F(s) representing a Laplace transform

There are two ways to represent this equation:

... or,

*ℒ - Laplace transform

The Laplace transform ℒ is a linear in the sense that ℒ[C1f1 + C2f2] = C1ℒ[f1] + C1ℒ[f2], where C1, C2 are constants.

Some more theory & example problems:

> Laplace Proofing #1

> Example 8.6

> Example 8.2

> Example 8.3

Diff. Eq. - finding particular solution using the Method of Undetermined Coefficients

 Given the equation:

y'' - 4y' + 3y = e^(3x) * (6+8x)

(a) Find a particular solution of the equation by using the Method of Undetermined Coefficients.

(b) Find the general solution of the equation.

(Start by finding the auxiliary equation of the given diff. equation)

y'' - 4y' + 3y = 0  →  r^2 - 4r + 3y = 0...

► r = 3, 1

► y = e^x & e^3x

Based off your general solution answers, find a particular solution that matches the f(x) to the original function: (e^(3x) * (6+8x))

* Remember that e^(3x) was one of the roots, and we should avoid creating two roots that are exactly the same.

Thus, multiply by 'x'.

yp = xe^3x * (A+Bx) or, e^3x * (Ax+Bx^2)

(Let's start differentiating to get y'p and y''p)

* differentiate by using product rule.

y'p = [e^3x * (Ax+Bx^2)'] + [e^3x' * (Ax+Bx^2)]

►  = e^3x * (A+3Ax+2Bx+3Bx^2)

y''p = [e^3x * (A+3Ax+2Bx+3Bx^2)'] + [e^3x' * (A+3Ax+2Bx+3Bx^2)]

►  = e^3x * (6A+2B+9Ax+12Bx+9Bx^2)

Now, plug in your values into the given differential equation you were first handed.

y'' - 4y' + 3y = e^(3x) * (6+8x)

► (e^3x * (6A+2B+9Ax+12Bx+9Bx^2)) - 4[e^3x * (A+3Ax+2Bx+3Bx^2)] + 3[e^3x * (A+3Ax+2Bx+3Bx^2)] = e^(3x) * (6+8x)

Cancel / divide term e^(3x)

► ((6A+2B+9Ax+12Bx+9Bx^2)) - 4[(A+3Ax+2Bx+3Bx^2)] + 3[(A+3Ax+2Bx+3Bx^2)] = (6+8x)

Now distribute and simplify (this is a very tedious process!)

2A + 2B + 4Bx = 6+8x

►  (2A + 2B) = 6

►  4B = 8

→ B = 2, A = 1

Blue in A & B values for the yp and make your general solution as well.

Answers are....

(a) yp = xe^(3x) * (1+2x)

(b) y = C1e^x + C2e^3x + e^3x * (x+2x^2)