Power Series Solutions of Differential Equations

Key Questions

  • How do you find a Power Series solution of a nonhomogeneous differential equation?

    Assuming you know how to find a power series solution for a linear differential equation around the point x_0, you just have to expand the source term into a Taylor series around x_0 and proceed as usual.

    This may add considerable effort to the solution and if the power series solution can be identified as an elementary function, it's generally easier to just solve the homogeneous equation and use either the method of undetermined coefficients or the method of variation of parameters.

    Vinicius M. G. Silveira · 1 · Dec 18 2014
  • How do you use Power Series to solve the differential equation y'-y=0 ?

    The solution is

    y=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x,

    where c_0 is any constant.

    Let us look at some details.

    Let
    y=sum_{n=0}^infty c_n x^n
    y'=sum_{n=1}^infty nc_n x^{n-1}=sum_{n=0}^infty(n+1)c_{n+1}x^n

    So, we can rewrite y'-y=0 as

    sum_{n=0}^infty (n+1)c_{n+1} x^n-sum_{n=0}^infty c_n x^n=0

    by combining the summations,

    Rightarrow sum_{n=0}^infty[(n+1)c_{n+1}-c_n]x^n=0

    so, we have

    (n+1)c_{n+1}-c_n=0 Rightarrow c_{n+1}=1/{n+1}c_n

    Let us observe the first few terms.

    c_1=1/1c_0=1/{1!}c_0

    c_2=1/2c_1=1/{2}cdot1/{1!}c_0=1/{2!}c_0

    c_3=1/3c_2=1/3cdot1/{2!}c_0=1/{3!}c_0
    .
    .
    .
    c_n=1/{n!}c_0

    Hence, the solution is

    y=sum_{n=0}^infty1/{n!}c_0x^n=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x,

    where c_0 is any constant.

    Wataru · 1 · Oct 5 2014
  • How do you use Power Series to solve the differential equation y''+2xy'+y=0 ?

    Answer:

    See below

    Explanation:

    Assuming a power series solution like this:

    • y = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ldots = sum_0^oo a_n x^n

    • implies y^' = sum_1^oo n a_n x^(n-1) qquad qquad y^('')= sum_2^oo n (n-1)a_n x^(n-2)

    With this power series:

    y''+2xy'+y = 0

    implies underbrace(sum_2^oo n (n-1)a_n x^(n-2))_(=sum_0^oo (m+2) (m+1)a_(m+2) x^(m)) + 2x sum_1^oo n a_n x^(n-1) + sum_0^oo a_n x^n= 0

    implies sum_0^oo (n+2) (n+1)a_(n+2) x^n + 2 sum_1^oo n a_n x^n + sum_0^oo a_n x^n= 0

    implies 2 a_2 + sum_1^oo (n+2) (n+1)a_(n+2) x^n + 2 sum_1^oo n a_n x^n + a_0 + sum_1^oo a_n x^n= 0

    :. underbrace(2 a_2 + a_0)_(=0) + sum_1^oo ( underbrace( (n+2) (n+1)a_(n+2) + (2 n +1)a_n )_(=0) )x^n = 0

    This insists that the coefficient of the x^0 term is zero, but also that the coefficient of every x^(i gt 0) term is also zero, as the homogeneous DE requires.

    The recurrence relation is:

    • a_(n+2) = - ((2 n +1))/((n+2) (n+1)) a_n

    This suggest 2 independent solutions to the DE, one for odd terms and one for even. These can be linearly super-imposed to reach a general null solution.

    Even terms:

    Setting: a_0 = 1, a_1 = 0:

    {( a_0 = 1),(a_2 = - 1/2 ),(a_4 = - 5/(4*3)* - 1/2 = 5/(4!) ),(a_6 = -(9)/(6*5) * 5/(4!) = - (5*9)/(6!) ),(a_8 = -(13)/(8*7) * - (5*9)/(6!) = (5*9*13)/(8!) ):}

    implies underbrace(1/(1!) x^0)_(k = 0) + underbrace((-1)^(1) 1/(2!) x^2)_(k = 1) + (-1)^(2) 5/(4!) x^4 + (-1)^(3) (5*9)/(6!) x^8 + (-1)^(4) (5*9*13)/(8!) x^8 + cdots + underbrace((-1)^(k) (5* 9 * ... * (4 k - 3))/((2k)!) x^(2 k))_("but not for first 2 terms") + cdots = 0

    An even solution is therefore:

    y_E = 1 - x^2/2 + sum_2^oo (-1)^(k) (5* 9 * ... * (4 k - 3))/((2k)!) x^(2 k)

    Odd terms:

    Setting: a_0 = 0, a_1 = 1, and copying the broad pattern:

    {( a_1 = 1),(a_3 =- 3/(2*3) = - 3/(3!) ),(a_5 = - 7/(5*4)* - 3/(3!) = (3*7)/(5!) ),(a_7 = - 11/(7*6) * (3*7)/(5!) =- (3*7*11)/(7!)),(a_9 = -(3*7*11)/(7!)* -(15)/((9*8) )= (3*7*11*15)/(9!) ):}

    implies underbrace( x )_(k = 0) + underbrace((-1)^(1) 3/(3!) x^3)_(k = 1) + (-1)^(2) (3*7)/(5!) x^5 + (-1)^(3) (3*7*11)/(7!) x^7 + cdots + underbrace((-1)^(k) (3* 7 * ... * (4 k -1))/((2k+1)!) x^((2 k+1)))_("but not for first term") + cdots = 0

    An odd solution is therefore:

    y_O =x + sum_1^oo (-1)^(k) (3* 7 * ... * (4 k -1))/((2k+1)!) x^((2 k+1))

    Recognising the linearity:

    y = c_1\ y_O + c_2 \ y_E

    So you have to add all that up

    Finally , a screen grab from Socratic that always puts me of answering qu's like this in proper fashion:

    Socratic

    I'd recommend: this

    Ultrilliam · 2 · Jul 23 2018
  • How do you find a Power Series solution of a differential equation?

    Let us solve the differential equation y''=y by Power Series Method.

    Let y=sum_{n=0}^inftyc_n x^n, where c_n is to be determined.
    By taking derivatives term by term,
    y'=sum_{n=1}^{infty}nc_nx^{n-1}
    and
    y''=sum_{n=2}^infty n(n-1)c_nx^{n-2}

    So, y''=y becomes
    sum_{n=2}^infty n(n-1)c_nx^{n-2}=sum_{n=0}^inftyc_n x^n
    by shifting the indices on the summation on the left by 2,
    sum_{n=0}^infty(n+2)(n+1)c_{n+2}x^n=sum_{n=0}^inftyc_n x^n

    By matching each coefficients,
    (n+2)(n+1)c_{n+2}=c_n Rightarrow c_{n+2}=c_n/{(n+2)(n+1)}

    Let us observe the first few even terms,
    c_2=1/{2cdot1}c_0=1/{2!}c_0
    c_4=1/{4cdot3}c_2=1/{4cdot3}cdot1/{2!}c_0=1/{4!}c_0
    .
    .
    .
    c_{2n}=1/{(2n)!}c_0

    Let us observe the first few odd terms,
    c_3=1/{3cdot2}c_1=1/{3!}c_1
    c_5=1/{5cdot4}c_3=1/{5cdot4}cdot1/{3!}c_1=1/{5!}c_1
    .
    .
    .
    c_{2n+1}=1/{(2n+1)!}c_1

    Now, we can find the solution y.
    y=sum_{n=0}^infty c_nx^n
    by separating even terms and odd terms,
    =sum_{n=0}^inftyc_{2n}x^{2n}+sum_{n=0}^inftyc_{2n+1}x^{2n+1}
    by using the formulas for c_{2n} and c_{2n+1} above,
    =c_0sum_{n=0}^inftyx^{2n}/{(2n)!}+c_1sum_{n=0}^infty x^{2n+1}/{(2n+1)!}

    Recall:
    coshx=sum_{n=0}^infty x^{2n}/{(2n)!}
    sinhx=sum_{n=0}^infty x^{2n+1}/{(2n+1)!}

    Hence, y=c_0coshx+c_1sinhx, where c_0 and c_1 are any constants.

    Wataru · · Sep 14 2014

Questions

Power Series