How do you use Power Series to solve the differential equation #y''+2xy'+y=0# ?

Answer 1

See below

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:

    I'd recommend: this

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Answer 2

To solve the differential equation ( y'' + 2xy' + y = 0 ) using power series, we assume that the solution can be represented by a power series:

[ y(x) = \sum_{n=0}^\infty a_nx^n ]

Substituting this series into the given differential equation and equating coefficients of like powers of ( x ) yields a recursion relation for the coefficients ( a_n ). By solving this recursion relation, we can find the coefficients of the power series solution.

First, differentiate ( y(x) ) twice and substitute into the differential equation. Then, equate coefficients of like powers of ( x ) to obtain the recursion relation. Solve this relation to find the coefficients ( a_n ).

Finally, the power series solution ( y(x) ) is the sum of the terms with the found coefficients ( a_n ), and it converges within a certain radius of convergence determined by the solution's behavior.

This method allows us to find a series representation for the solution of the given differential equation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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