Power series solution of differential equations Puns

Hi guys,

I am learning about using differential equations to find power series solutions and I have come across a question I can find any information on how to handle.

The Question: Find a power series solution of the following differential equation.

y' - x - y = 0

With y(0) = 0 and the equation is about the point X0 = 0

The part I am stuck on is what happens to the -x in my equation, since I know the series for y(x) and y'(x) but I can't find any information on what happens with the -x , I thought about differentiating it once more but it doesn't give a value for y'(0) so I didnt go ahead with it

https://imgur.com/G7RQHa4

Problem: Find the power series solution x = Σ(from n=0 to infinity) a_n*t^n of the differential equation x" = t^(2)x, which satisfies the initial conditions a_0 = 1 and a_1 = 2.

I have done most of the work here: http://imgur.com/v79eGU3

I have found that a_2 and a_3 are both equal to zero, as well as what a_n is equal to. I'm just not sure how to finish the problem so that I can write the solution of x as a power series.

Thanks in advance, and please let me know if you can't read my handwriting!

https://media.cheggcdn.com/study/667/66701bd0-c313-4f14-8773-db928e32c138/image.png

Need help with a and b, thanks in advance

I feel like the question has an error and the equation should end with +y instead of -y

The problem is as follows:

Show that the general solution to x^2y''-(3/2)xy'+(1+x)y=0 can be expressed as

y = C1*x^2*Sum(k=0 to infinity)((-1)^k*(k+1)*2^(2k)*x^k)/(2k+3)! + C2*x^(1/2)*(1+2x+(1/4)Sum(k=2 to infinity)((-1)^k*(k+1)*2^(2k)*x^k)/(k(k-1)(2k-3)!)

I have found a recurrence relation for the first summation, where ak = (-1)^k/(k!(k+2/3). Multiplying by 2 gives (-1)^k*2/(k!(2k+3), but does not get me to the proper summation. I have no idea where that factor of k+1 is coming from, or why the 2 should be raised to the 2k power. I think getting help with the first summation will also help me figure the second one out. Please and thanks.

http://imgur.com/7J1qOY4 Can anyone explain the method used to solve the problem above? I am only familiar with the more common method of solving power series differential equations that involves setting y equal to a generic power series, plugging that and its derivatives in, and then using a recurrence relation one it's all consolidated into one power series. Any help is appreciated!

We weren't taught this material during the course due to time restrictions. We were given an online assignment to be completed over the break on it anyway. Can someone show me how to solve

1.y''−(sinx)y=cosx, y(0)=9 y'(0)=−6 2.(x^(2)−x+1)y''−y'+5y=0, y(0)=0,y'(0)=4

That would be great, thanks!

Edit: only number 2!!

Solve the initial value problem

y'' + 2y' + (1 - t/3)y = 0, y(0) = 1, y'(0) = 0

Find at least the first four nonzero terms in a power series expansion about t = 0 for y(t).

I really need the most help with finding the unique solution and the first four nonzero terms. How do I solve for the constants in the general solution? I know I need to derive the series, but I'm unsure of how to do that. In this case, will I increment the indices when I derive the series? We don't do that when we are say solving for the recurrence relationship.

If I end up with two power series in my unique solution, what is included in one nonzero term? Is it the 0th indices of both series combined or is the first nonzero term the 0th index of one and the second nonzero term is the 0th index of the other?

Find at least the first four nonzero terms in a power series expansion about x=0 for a general solution to the given differential equation.

(2x+3)y"-xy'+y=0

So after I get it to standard form by dividing by 2x+3, a point of singularity arises at -3/2. I just don't know what to do with this information now. Please help me, any tidbit will be appreciated.

http://i.imgur.com/LsWUa.jpg is the problem and my solution so far. the recurrence relation is not actually required and is extra credit. I'm fairly confident I am correct up to the point where I assume a0=c1 and a1=c2. Am I correct in this assumption and am I taking the right path after that assumption? I am struggling because we have not yet solved this type of problem without knowing the recurrence relation.

I'm beginning to study partial differential equations and I randomly thought solving this problem would be interesting but I can't find a solution online.

Hey everyone, so I already know how to use MATLAB to solve first, second and third order ode (with and without initial conditions) but I can’t seem to find a way to find a series solution to an ode. Is it even possible?

Hello all,

I've come across something I can't quite wrap my head around. when solving a DE using the Frobenius method I find the r values of the indicial equation. Here's what I don't get. For said example I had

r= -1/2 and r = 1

I've been told to use the higher value number as the lower might not work out (we aren't incorporating the relation to logs in our class but I've read that's where the lower value for r can sometimes come into play). I've also seen my teacher choose to use the lower number in this example.

Here's the problem: ​

2x^(2)y''+xy'-(1+2x^(2))y=0

I don't understand the logic behind picking one or the other enough to make a calculated decision.

I have that the two solutions are

[1] and [2]

and due to the linearity principle, also

[3]

for a linear system of two differential equations from my textbook. It doesn't explain why though. Can someone point a resource to me that can explain it in simple-ish terms? Where does the constant e come from?

Example problem from a quiz our instructor gave us:

y'' + xy' + y = 0 With x_0 = 0. For each of the functions in the fundamental set of solutions, display at least 4 terms.

We have been on this topic for a couple of weeks now and I was hoping I would get a hold on it, but I haven't. I honestly don't understand anything about it.

If it helps, we are using Elementary Differential Equations and Boundary Value Problems by Boyce and DiPrima. It's chapter 5 of that book.

I am trying to solve this D.E: y'-xy=0

i know that:

y=Σan*x^n and y'=Σ(n+1)*an+1*x^n

So y'-xy=0 =>y'=xy=> Σn*an*x^n=Σan*x^n+1

So we get the following equation:

(n+1)an+1*x^n=an*x^n+1 => an=((n+1)an+1)/x

for n=0: ao=a1/x

for n=1: a1=2*a2/x

for n=2: a2=3*a3/x

...

an=(n+1)an+1/x

Let a0=c, so we try to find a retroactive equation in order to get the recurrence relation:

for n=0: a0=c

for n=1: a1=a0x=cx

for n=2: a2=a1x/2=(cx^2)/2

for n=3: a3=a2x/3=(cx^3)/6

...

So i find that an=(cx^n)/(n+1)!

But the correct answer is an=(cx^n)/(n!2^n) so at what point i am wrong?

Problem: https://imgur.com/xoAmazY

I for the life of me don't know where to start. I have no idea what v and x stands for (im guessing it is for velocity and distance???). Can anyone give me hints on how what it means or how to start it (no answers as usual). I have no problem with answering simpler problems like this one, https://imgur.com/A9KYUqN, but have no clue with the former.