Inverse Laplace Transform and Court

[m_celikok]

Homework Statement

I had a question in my midterm, it was to find inverse laplace tansform of:
(4s+5) / (s^2 + 5s + 18.5)

Where ^ denotes power.

Homework Equations

The Attempt at a Solution

My answer was to find the complex roots of equation (s^2 + 5s + 18.5) , by them transform it to form (s - w0)(s - w1) where w0 and w1 are complex roots of original equation. Then I divided the total rational (4s+5) / (s^2 + 5s + 18.5) into sum of two different rationals in form of (A)/(s -w0) + (B)/(s-w1). Found A and B and then solved each sum as different inverse laplaces. My solution was,

Yet my teacher insists this is not the solution and gave me a 0 from that question. I am going to take this to Court. Can you confirm my solution please?

 

[pasmith]

Homework Statement

I had a question in my midterm, it was to find inverse laplace tansform of:
(4s+5) / (s^2 + 5s + 18.5)

Where ^ denotes power.

Homework Equations

The Attempt at a Solution

My answer was to find the complex roots of equation (s^2 + 5s + 18.5) , by them transform it to form (s - w0)(s - w1) where w0 and w1 are complex roots of original equation. Then I divided the total rational (4s+5) / (s^2 + 5s + 18.5) into sum of two different rationals in form of (A)/(s -w0) + (B)/(s-w1). Found A and B and then solved each sum as different inverse laplaces. My solution was,

Yet my teacher insists this is not the solution and gave me a 0 from that question. I am going to take this to Court. Can you confirm my solution please?

Your solution is approximately equal to <br /> 4e^{-2.5t}\cos(3.5 t) - \tfrac{10}{7} e^{-2.5t}\sin(3.5 t)<br /> which can be obtained by completing the square in the denominator of the transform and using a table of transforms.

You should get some credit for having found a substantially correct answer by a correct method, but you really need to express the coefficients as exact fractions, not decimal approximations, and express the complex exponentials in terms of cosines and sines.

 

[m_celikok]

fractions

the coefficients are expressed as fractions in my paper i just calculated them from wolfram alpha to easily post here. also i noted at the end of my solution that e's in this solution can be expressed as cosine and sines through euler identity but i rathered not which is because i find this form clearer. this was an computer engineering , systems and signals course so no such rigor was required. my teacher is quite hostile towards me. thanks for your reply sir.

 

[HallsofIvy]

I would NOT try to factor the denominator. Instead, I would complete the square:
s^2+ 5s+ 18.5= s^2+ 5s+ \frac{25}{4}- \frac{25}{4}+ \frac{74}{5}= (s+ \frac{5}{2})^2+ \frac{49}{4}

So \frac{4s+ 5}{s^2+ 5s+ 18.5}= 4\frac{s+ 5/2}{(s+ 5/2)^2+ (7/2)^2}+ \frac{10}{7}\frac{7/2}{(s+ 5/2)^2+ (7/2)^2}
Which can be looked up in a table of transforms.

 

[m_celikok]

I would NOT try to factor the denominator. Instead, I would complete the square:
s^2+ 5s+ 18.5= s^2+ 5s+ \frac{25}{4}- \frac{25}{4}+ \frac{74}{5}= (s+ \frac{5}{2})^2+ \frac{49}{4}

So \frac{4s+ 5}{s^2+ 5s+ 18.5}= 4\frac{s+ 5/2}{(s+ 5/2)^2+ (7/2)^2}+ \frac{10}{7}\frac{7/2}{(s+ 5/2)^2+ (7/2)^2}
Which can be looked up in a table of transforms.

Thanks for the reply. On the other hand the question here is not a "wouldn't". It is a "can/can't" question. Table way might be easier and cleaner which i agree too. But, does that invalidates my solution?