Please substitute y = x^2 to your differential equation, and you will find that it is indeed one of the solution.
y = x^2
y' = 2x
y'' = 2
Therefore:
x^2 y'' -3x y' + 4y = 2 x^2 - 6 x^2 + 4x^2 = 0
Hello. psholtz. I thought you just do the integration without considering their dependence. It is interesting that x\sqrt{1-y^2} + y\sqrt{1-x^2} = C is a solution. Thank you for your information in the Jacobian as well.
In fact, what you have show is that, x^2 is one of the solution. Since for all other terms other than a_0, they have to be zero in order for the expression to equal to zero.
I am sorry, but your 2nd equation:
\sqrt{1-y^2}dx + \sqrt{1-x^2}dy = 0
cannot be integrated to get the 3rd equation:
x\sqrt{1-y^2} + y\sqrt{1-x^2} = C
You cannot integrate term by term, from dx to x, since \sqrt{1-y^2} depends on x, and \sqrt{1-x^2} depends on y.
In fact, if you...
Hello. Please refer to my article in http://www.voofie.com/concept/Mathematics/" [Broken]:
http://www.voofie.com/content/18/solving-system-of-first-order-linear-differential-equations-with-matrix-exponential-method/" [Broken]
What you really need is matrix exponential, instead of matrix...
For your first equation, please refer to this question in http://www.voofie.com/concept/Mathematics/" [Broken]:
http://www.voofie.com/content/152/how-to-prove-eat-e-at_0-eat-t_0/" [Broken]
I think you typed wrong in this formula:
exp(At)_t=0 = I
0 is not equal to I. And your what's your...
Hello ferry2, I have solved your 2nd equation here:
http://www.voofie.com/content/146/how-to-solve-2x12-y--2-2x1-y-4-y-0/" [Broken]
And the solution is given by:
y(x) = C_1(2x+1) +C_2 (2x+1) \ln (2x+1)
Hello, beetle2. It didn't mentioned 2nd order with variable coefficients, since there is no general method in solving that. However, what you can do is by guessing the first solution. Then you can find the 2nd one using various method, such as...
If it is not a must that you have to use method of undetermined coefficients, you can have a look of the operator method. In this case, you don't have to care about what kind of non-homogeneous function you have. At least you can write the solution in integral form. Please refer to my tutorial...
You didn't do the shifting of the summation variables. You should do the following:
\sum _{k=0}^{\infty } \frac{(-1)^k(2k+1)(2k)x^{2k+1}+(-1)^k(2k+1)x^{2k+1}+(-1)^kx^{2k+3}-(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}
\sum _{k=0}^{\infty }...
You should have expanded the term:
\left(x^2-1\right)\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}
into 2 terms:
x^2\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}-\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}
You shift the index, and...
First of all, for the 2nd order differential equation, the Wikipedia method is correct, but not easy to understand or use. Normally we would write:
\begin{align*} \frac{d^2y}{dt^2}&=\frac{dy'}{dt} \\ &=\frac{dy'}{dy} \frac{dy}{dt}\\ &= y' \frac{dy'}{dy} \end{align*}
And the general method...
Your 1st solution of y = x is correct, but the other solution is wrong.
Please refer to the below question in http://www.voofie.com" [Broken].
http://www.voofie.com/content/86/how-to-solve-this-2nd-order-linear-differential-equation/" [Broken]
The question solves an inhomogeneous...
It is not you know all solutions. You know one solution of the homogeneous equation of order n, and can reduce the equation to n-1 order. And now, you just need to find a solution to the n-1 order equation. It may be simpler, but it may be more difficult too.
@jasonRF,
You may say my method...
Here is a paper written by me:
http://www.voofie.com/content/84/solving-linear-non-homogeneous-ordinary-differential-equation-with-variable-coefficients-with-operat/" [Broken]
Basically, it uses operator method to solve linear non-homogeneous ordinary differential equation with variable...
I am sorry. I have little bit of typo in the answer.
It should be like this:
\Rightarrow b_k=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(e^{\frac{2k \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)}
You don't understand why there is a factor of (-a)^{\frac{N-1}{N}}...
I am sorry. I think my notation confused you.
I wrote this as the equation:
\sum _{k=0}^{N-1} b_k\prod _{j=0j\neq k}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)=1
But in fact, i really means this:
\sum...
@Gregg,
Have you try putting x = \sqrt[N]{A}e^{\frac{\theta +2n \pi }{N}i}?
If you do that, you can see for every other term in the sum, i.e. when k \neq n
b_k\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(\sqrt[N]{A}e^{\frac{\theta +2n \pi }{N}i}-\sqrt[N]{A}e^{\frac{\theta +2j \pi...
Actually, this integral can be computer if N is integer, without using hypergeometric function.
\int \frac{dx}{x^N+ a}=\sum _{k=0}^{N-1} b_k\ln \left(x-(-a)^{\frac{1}{N}}e^{\frac{2k \pi }{N}i}\right)
and
b_n=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{\underset{j\neq k}{j=0}}^{N-1}...
I am so glad that you read my tutorial and solved the problem yourself. It is rather strange that you can do a partial fraction on the differential operators. But I think it is just due to the algebraic structure of the differential operator.
You can try powers of x for the inhomogeneous function of 1.
Ax^2+Bx+C
If your exam allows, you may try the operator method. It works for all inhomogeneous function. You may refer to my tutorial in http://www.voofie.com" [Broken]...
I am now providing a method to reduce the equation into system of 1st-order differential equations, so that you can use matrix exponential to solve it easily.
\mathbf{x}'= C \mathbf{y}
\mathbf{y}'=\mathbf{x}
\frac{d}{\text{dt}}\left(
\begin{array}{c}
\mathbf{x} \\
\mathbf{y}...
Have you tried using operator method? It can deal with linear differential equation with arbitrary inhomogeneous function easily.
Here is a reference for the method...