Differential Equations: Non-homogeneous Series Expansion

Bryon
Messages
98
Reaction score
0

Homework Statement



y'' + y' + y = 1 + x + x2

Homework Equations



y = Ʃ CN*xN N starts at 0
y' = Ʃ N*CN*x(N-1) N starts at 1
y'' = Ʃ N*(N-1)*CN*x(N-2) N starts at 2

3. The Attempt at a Solution [/]
I know how solve the equations using series when the equation would equal to 0. My main question about using series on a non-homogeneous differential equation is whether or not the varialbes on the right side have the Cx coefficients? Or would they be paired up with the x, x2, etc? I think I need some quick clarification on this.

Thanks!
 
Physics news on Phys.org
Bryon said:

Homework Statement



y'' + y' + y = 1 + x + x2

Homework Equations



y = Ʃ CN*xN N starts at 0
y' = Ʃ N*CN*x(N-1) N starts at 1
y'' = Ʃ N*(N-1)*CN*x(N-2) N starts at 2

3. The Attempt at a Solution [/]
I know how solve the equations using series when the equation would equal to 0. My main question about using series on a non-homogeneous differential equation is whether or not the varialbes on the right side have the Cx coefficients? Or would they be paired up with the x, x2, etc? I think I need some quick clarification on this.

Thanks!


The Cn's only appear in your expressions for y and its derivatives. But you must take the powers of x on the other side into account for your recursion formulas. I assume you know that series isn't the easiest way for this problem.
 
Thanks for clearing that up. The instructor covered only homogenous problems, and when I ran into one of these I was not entirely sure how to solve it with series.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top