Recent content by rey242

Homework Statement A voltmeter will be used to measure the voltage vA in the circuit below. Assume that the voltmeter has a resistance of 100[ohms] and a full-scale voltage of 20[V]. a) What will the voltmeter read? b) What is the percentage error in making this reading?Homework Equations...

What procedure would I use to find a and b? I know about the error...would I convert the sum of the errors squared into normal equations?

Homework Statement For the following data, find the least squares fit of the given form x=1,2,3,4,6 y=14,10,8,6,5 h(x)=ae^x+be^(-x) Homework Equations The Attempt at a Solution So I tried to linearize the equation by taking the natural log of everything...

Such as a vector U(0)=(1;1)

so what if it included a initial solution, would that change anything?

I wrote the question a little wrong... I meant U'(t)=A u(t)+V Where A is just a regular 2X2 matrix and V is a 2x1 vector. like this U'(t)=(1,2;3,4)U(t)+(1;e^t) I already know how to solve systems such as U'(t)=Au(t) by using the eigenvalues & eigenvectors of A but I haven't learned...

Hey guys, I have a quick question about adding a vector to a system of differential equations. Like U'(t)= Ax+V Say A is a 2x2 matrix and V is a 2x1 vector. Can you all explain how I could handle that V vector? Should somehow include it into A?

Remember Integration by Parts has "tricks"? Try setting your equation to \int exp(t/10)sin(t)dt =20et/10 - et/10cos(t) - (1/10)et/10sin(t) - (1/100)\int et/10sin(t)dt

Yeah sorry about that, I forgot. :eek: Multiply both sides by exp(.1t). The integration is harder now but its still the basic concept.

Thanks for all your help I understand now. I really appreciate it!

Try my way... e^(.1t)x'+.1e^(.1t)x=2+sin(t) The left side is just the product rule. So if you do this: d/dt(e^(.1t)*x)=2+sin(t) Then you can do a really simple integration. But you could also do it Dick's way. :)

The multiplicative inverse is 1/(x-3), this does not apply when x=3... Hmm... I was thinking about this before but I felt it wasn't right. Should I multiply row 3 by the multiplicative inverse ?

Here is my process: 3, 0, 6, 1, 0, 0; 1, -2, x, 0, 1, 0; 1, 2, 1, 0, 0, 1; I multiplied -1/3 to r1 and added to r2 and got 3, 0, 6, 1, 0, 0; 0, -2, x-2, -1/3, 1, 0; 1, 2, 1, 0, 0, 1; I then did the same and added to r3 and got 3, 0, 6, 1, 0, 0; 0, -2, x-2, -1/3, 1, 0; 0, 2, -1, -1/3, 0, 1; I...

You're on the right track. Remember you have a dx/dt on the left so the equation would be (dx/dt)+.1x= 2+sin(t) by turning this into a different notation you get: x'+.1x=2+sin(t) So then you can now multiply by the integrating factor and get e^(.1t)x'+.1e^(.1t)x=2+sin(t) Does the left side...

Ok, I reworked it and I got this: 3, 0, 6, 1, 0, 0; 0, -2, 1, 1/3, 0, -1; 0, 0, (X-3), -2/3, 1, 1; So would this matrix not have an inverse since I can never get the Identity on the left? It seems like it I'll keep getting number when I do row reduction after this...