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)

Statics - Fundamental Problem 7.6

   
From Chapter 7, 

Assume A is pinned and B is a roller in (Figure 1). Take that w = 4.5 kN/m . Determine the normal force, shear force, and moment at point C. Follow the sign convention.

▼ Three Part Question:

Determine the normal force, shear force, and moment at point C. Express your answer to three significant figures and include the appropriate units.

Step 1 - always make your free body diagram! In the instructions given, point A is a pinned support and point B is a roller. The triangular and distributed load need to be expressed as a concetrated point load as well:

Make your free body diagram! In the instructions given, point A is a pinned support and point B is a roller. The triangular and distributed load need to be expressed as a concetrated point load as well.

Substituting 4.5 for w, We get that By =