# Using laplace transforms to solve IVP problems with cramers rule

• fufufu
In summary, we are given the differential equation y'' + 4y = sin(2t), with initial conditions y(0) = 1, y'(0) = 1. After isolating Y(s), we get Y(s) = 2 / (s^2+4)^2 + s/(s^2+4). Using partial fraction decomposition and solving for the coefficients, we obtain A=B=C=0 and D=2. This simplifies Y(s) to 2/(s^2+4)^2 + 2/(s^2+4), which can be further simplified to 2sin(2t) in the time domain. However, the book solution is cos(2t
fufufu

## Homework Statement

y'' + 4y = sin(2t)

y(0) = 1
y'(0) = 1

cramers rule

## The Attempt at a Solution

after isolating Y(s) i end up with:

Y(s) = 2 / (s^2+4)^2 + s/(s^2+4)

after i do partial fraction decomposition, i get 4 equations with 4 unknowns, A, B, C and D

i inserted B = 2- 4D into other 3 equations..

when i used cramers rule to get values for A , B and C I am getting zeroes for all three variables. (??) ..also, I noticed that each equation also equals zero, but I am not seeing what that might mean?
i am doing something wrong

my 3 equations are:

A + 5C + D = 0
-A -5C + D = 0
2A + 16C -4D = 0

any help is really appreciated...thanks

## The Attempt at a Solution

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I don't get the same thing as you did for Y(s). Are both y(0) and y'(0) supposed to equal 1, or is one of them equal to 0?

It would help if you showed what terms you're trying to break Y(s) into.

Thanks for the reply. yup you're right i copied down one of the IV's wrong; should be y(0) = 1 and y'(0) = 0. But I actually solved with the correct IV's, so my Y(s) is the same:
heres how i got it

s^2Y(s) -sf(0) - f'(0) + 4Y(s) = 2/(s^+4)

Y(s) = 2 / (s^2+4)^2 + s/(s^2+4)

Y(s) = 2/(s^2+4)^2 + 2/(s^2+4)

i decomposed just the first term on RHS into this

2/(s^2+4)^2 = As+B/(s^2+4) + Cs+D/(s^2+4)^2

multiplying by the denom i get:

2 = As+B + Cs+D/(s^2+4)

i next plugged in values of s=-1, 0, 1 and 2

this gave me 4 equations and 4 unknowns

i then solved for B in terms of D which is B=2-4D

i am now left with 3 equations 3 unknowns

A + 5C + D = 0
-A -5C + D = 0
2A + 16C + 4D = 0but like i said in initial post, i am puzzled as to what all the zeroes mean? (equations wrong somehow?)

Also, a quick aside question: i just learned cramers rule for 3 equations and 3 unknowns does the rule work exactly the same for 4 equations with 4 unknowns? (and beyond?) and also, the method seems really susceptible to basic book keeping mistakes.. is this just because i just learned it, or is it not a very fast and efficient method for solving equations? thanks again

UPDATE: i just plugged in my (originally) 4 equations into cramers rule calculator and got A=C=D=0 and B=2

that would mean my decomposition = 2/s^2+4 + 2/s^2+4

= sin2t + sin2t = 2sin2t
but the book solution is cos2t - (1/4)tcos2t + (1/8) sin2t

Last edited:
fufufu said:
i decomposed just the first term on RHS into this

2/(s^2+4)^2 = As+B/(s^2+4) + Cs+D/(s^2+4)^2
You really need to get in the habit of using parentheses. You might notice that the LHS is exactly what you'd get when A=B=C=0 and D=2. In other words, it's already in the form you want.

multiplying by the denom i get:

2 = As+B + Cs+D/(s^2+4)
You made an algebra mistake. Since you multiplied by (s2+4)2, you should have
$$2 = (As+B)(s^2+4) + Cs+D$$ If you solve for the coefficients, you'll indeed find that A=B=C=0 and D=2.
Also, a quick aside question: i just learned cramers rule for 3 equations and 3 unknowns does the rule work exactly the same for 4 equations with 4 unknowns? (and beyond?) and also, the method seems really susceptible to basic book keeping mistakes.. is this just because i just learned it, or is it not a very fast and efficient method for solving equations? thanks again
Yes, Cramer's rule works for any number of unknowns. I find it's actually a nice method to use when you have unpleasant coefficients. For larger systems, it gets unwieldy simply because calculating determinants of large matrices is a pain. The largest system I'd use it for is probably one with three unknowns, unless I happened to have a lot of zero coefficients to make the big determinants easy to calculate.

UPDATE: i just plugged in my (originally) 4 equations into cramers rule calculator and got A=C=D=0 and B=2

that would mean my decomposition = 2/s^2+4 + 2/s^2+4

= sin2t + sin2t = 2sin2t
but the book solution is cos2t - (1/4)tcos2t + (1/8) sin2t
You will find it helpful to consider the derivative
$$\frac{d}{ds} \left( \frac{2}{s^2+4} \right)$$ and use the property of the Laplace transform involving F'(s).

## What is a Laplace transform?

A Laplace transform is a mathematical technique used to convert a function from the time domain to the frequency domain. It is often used in engineering and physics to solve differential equations and systems of equations.

## How does Cramer's rule work?

Cramer's rule is a method of solving systems of linear equations by using determinants. It involves finding the determinants of the coefficient matrix and the augmented matrix, and then dividing them to find the values of the variables in the system.

## Can Laplace transforms be used to solve initial value problems?

Yes, Laplace transforms can be used to solve initial value problems. The Laplace transform can be applied to both sides of a differential equation to convert it into an algebraic equation, which can then be solved using Cramer's rule.

## What are the advantages of using Laplace transforms to solve IVP problems?

Using Laplace transforms to solve IVP problems can often be faster and more efficient than traditional methods. It also allows for the combination of multiple equations into one equation, making it easier to solve complex systems of equations.

## Are there any limitations to using Laplace transforms and Cramer's rule?

Yes, there are some limitations to using Laplace transforms and Cramer's rule. They are not always applicable to all types of differential equations and systems of equations. Additionally, the calculations can become more complex and time-consuming for larger systems.

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