Thread: Problems

Howdy!

A little introduction, my name is Alfonso, I'm a Physics major. I'm orginally from Mexico but I've been living in the States for about half my life. I'm beginning my third year of college at Texas Tech and am thoroughly enjoying my time here! I'm actually a bit behind on my classes because I was transferred late and missed two days of school. This actually led me to my debut here! (I don't know if that's a good thing yet) My professor kindly lended my $200 to purchase my Differential Equations book for ODE about half an hour ago. I've been struggling with this homework. So far I've shot three problems down but the rest of these (posted) are perplexing me. I was wondering if I could get some help? I look forward to contributing in the future! =D


Problem #1
Find the position of a moving particle with the given acceleration , initial position , and initial velocity

.

____

Problem #2
The general solution to can be written in the form where is an arbitrary constant.

____ For this problem I keep getting (1/3) ln |y| = ln |2+x| +C Is this correct? Am I missing a step?

Problem #4
The solution to with is

____ For this problem I get y = square root of 2e^x + 2C is this correct?

Problem #5
The general solution of the first order linear equation can be written as where is an arbitrary constant.

____
Problem #6
Solve the first order linear Initial value problem with .

____
Problem #7
Solve the first order linear Initial value problem with .

____

Problem #8
Solve the equation by first setting to obtain a first order equation for . After you solve for recall that and solve the resulting pure time equation for .

The solution can be written as where and are arbitrary constants.

____

#1: x = x0 + vt + 1/2 at^2 = 2 - 10t + 1/2 * t^2 * (6t+2)

#2: I remember you've to separate the variables, i.e. y'/y = 3/(2+x) then integrate both sides.

#4: Put y = ae^(bx) then y' = abe^(bx) so yy' = a^2 b e^(2bx) = e^x which means b = 0.5 and a = +- sqrt(0.5). y(0) = 1 seems to be redundant info!

Originally Posted by knighttof3

#1: x = x0 + vt + 1/2 at^2 = 2 - 10t + 1/2 * t^2 * (6t+2)

#2: I remember you've to separate the variables, i.e. y'/y = 3/(2+x) then integrate both sides.

#4: Put y = ae^(bx) then y' = abe^(bx) so yy' = a^2 b e^(2bx) = e^x which means b = 0.5 and a = +- sqrt(0.5). y(0) = 1 seems to be redundant info!

#1 I'm not sure I follow, how did you reach this?
#2 That's what I did but I get stuck at the very end:
ln |y| = 3 ln |2+x| +C
I'm not sure how to get the natural log out of the left-hand side. I believe it's by applying e but I do not know how that affects the right-hand side. Also I'd like someone to confirm that.
#4 I have no idea what you did.

For #1:


In other words you integrate it once, find the constant, and integrate it again and find the constant.


For #2: what you have so far is right, you just have to rewrite it so you have

y= ________

So yes you raise both sides to an e power.



for #4:



but generally we write the 2C as just C because C can be anything. So it's writen as:

since your starting condition is y(0)=1, you need to solve for C.

For the others I'll have to find my textbook, I haven't done this in a while.

Hi, I'm actually probably in the same class as you if there's only one section. I had a problem with number 8 that I've now figured out but searching brought me to this page. I'll try and help as best I can.

Problem 1: The integral of the acceleration is the velocity, so first integrate a(t) and find C with the given v(0). The position is the integral of the velocity so integrate your new v(t) and plug in x(0) to answer the question.

Problem 2: You're on the right track, just raise e to both sides revealing y = e^3ln(x+2) which becomes (x+2)^3.

Problem 4: This is a separable eqn. Gather like terms and you come up with (y)dy = (e^x)dx. Integrate both sides, simplify and plug in your initial value to get your answer.

Problem 5: Subtract 2xy to the other side so it looks like the equation in Problem 6. Then it's a linear equation with Case 3 that we talked about in class. P(x) = 2x, Q(x) = e^x^2. μ(x) = e^integral(P(x)dx). (μ(x)y)' = μ(x)Q(x). Now you just substitute and solve.

Problem 6: Same as Problem 5 except you have a constant C that you have to find by plugging in the initial value.

Problem 7: Same concept as Problem 6, just start by dividing everything by x.

Problem 8: v=y' so your first equation becomes v' + v = 3x. Use the same way of solving it as the last 3 to get v which is equal to y' so just take the integral and you have your y.

Good luck, I guess I might see you in class tomorrow if we're in the same one.

Adam Particka