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 = 

Statics - Problem 6.15

  

From Chapter 6, Structural Analysis

In (Figure 1), a = 10 ft.$\mathrm{<span\; style="color:\; \#333333;\; font-family:\; arial;"> Members\; AB\; and\; BC </span>}$can each support a maximum compressive force of 950 lb, and members AD, DC, and BD can support a maximum tensile force of 

▼ PART A

Determine the greatest load  P the truss can support.

Express your answer to three significant figures and include the appropriate units.

 STEP 1  

Start by taking calculating the reaction forces about both support points.

Take moment abt. point A (which is a pin support.)

Solve for Cy;

Now, equate all vertical forces.

We have already solved for Cy, plug in its value;

 STEP 2 

Now, Create your FBD for the truss surrounding joint A.

Force AB is the force in member AB

& Force AD is the force in member AD.

Calculate the angles subtended by BAx & DAx.

BAx calculation:

*SOH CAH TOA

Use tangent for this angle calculation since we know the height (a) and the base (a).

DAx calculation:

*SOH CAH TOA

Use tangent for this angle calculation since we know the height (a/4) and the base (a).

 STEP 3 

Now, equate all horizontal forces.

Equate the forces along the vertical direction.

= 0.647P (tension)

 STEP 4 

Calculate the force in member AB:

= 0.943P (compression)

Due to symmetrical shape of ABD and BDC, the tensions in both members are equivalent.

= 0.647P

= 0.943P

Statics - Problem 4.47

 From Chapter 4, Force System Resultants

The cable exerts a 130-N force on the telephone pole shown in (Figure 1)

Determine the moment of this force about point A. Enter the components of the moment separated by commas. Express your answer in newton-meters to three significant figures.

 STEP 1  

Gather all presented data, as you will use it to your advantage in vector-type problems. Firstly, gather co-ordinates in i,j,k format of all the presented points.

→ A (0,0,0)

→ B (

Statics - Problem 2.24

 Statics, Chapter 2 - Force Vectors

In (Figure 1), the resultant F_R of the two forces acting on the jet aircraft should be directed along the positive x axis and have a magnitude of 11 kN,

▼ PART A

Determine the angle θ of the cable attached to the truck at B so that F_B is a minimum. Express your answer in degrees to three significant figures.

▼ PART B

What is the magnitude of force in cable  B when this occurs? Express your answer to three significant figures and include the appropriate units.

▼ PART C

What is the magnitude of force in cable  C when this occurs? Express your answer to three significant figures and include the appropriate units.

F_R = resultant, which is directed along the x-axis and = 11 kN.

θ = ? to have to be minimum...

since F_y = 0, F_c sin(20) = F_b sin(θ)

F_x = F_R = F_c cos(20) + F_b cos(θ) = 11 kN

Let's solve for F_c using the equation for Fy

Now simply the 1/sin(20) and plug F_c  in the equation for F_R

Now, take the derivative of 2.92sin(θ) + cos(θ) in order to get the maximum value that compliments the equation

Set the right hand side = 0, and solve the equation.

Divide every term by cos(θ).

Using trigonometric identities, evaluate -sin(θ)/cos(θ) = -tan(θ)

Take the inverse tan of both sides; arctan(2.92) = 71.1

Part A answ = 71.1°

Now plug in 71.1° in this equation from above...

... to get our value for F_b or F_b(min)

when doing so, our Part B answer is, F_b = 11/3.09 or 3.56 kN

Now for Part C, we can plug in what we now know to find F_c re-using the equation below...

F_c = 9.84 kN