differential equation, Math help

Question

Find the general solution of the equation by the power series method.

y''-2xy'-2y=x

Answer 1

This is a non-homogeneous equation. That means that its solution can be written as:

y_h + y_p

where y_h is the solution to the homogeneous equation

y'' -2xy' - 2y = 0.

Any y_p is a particular solution unique to the non-homogeneous equation. The particular solution is easier to find using the method of undetermined coefficients. Since the non-homogeneous term is linear, assume the particular solution is of the form

y_p = Ax+B

If this assumption does not work, you can consider higher order assumptions (quadratic, cubic etc...)

Plug this into the equation to give

y'' -2xy' - 2y = x => (Ax+B)'' - 2x(Ax+B)' -2(Ax+B) = 0

Evaluating the derivatives of the particular sulution:

y_p = Ax+B

y_p' = A

y_p'' = 0

=> (Ax+B)'' - 2x(Ax+B)' -2(Ax+B) = 0 => 0 -2x(A) - 2(Ax+B) = x => -2Ax -2Ax -2B = x

Setting up to compare coefficients gives

(-4A)x -2B = 1x + 0

-4A = 1

-2B = 0

This means that A = -1/4, and B=0, so y_p = (-x/4)

Next, consider the homogeneous equation

y'' -2xy' -2y = 0

Start with your solution y as a power series

Let ∑(k=b) a_k be the sum from k=b to k= infinity of the terms a_k

y=∑(k=0) a_k x^k

Term by term differentiation gives

y'=∑(k=1) k*a_k x^(k-1)

y''=∑(k=2) (k-1)*k*a_k x^(k-2)

Substituting into the differential equation gives

y''-2xy'-2y=x

∑(k=2) (k-1)*k*a_k x^(k-2) -2x(∑(k=1) k*a_k x^(k-1)) -2(∑(k=0) a_k x^k) = 0

Bringing the x and constant multipliers into the sums gives

∑(k=2) (k-1)*k*a_k x^(k-2) + (∑(k=1) -2*k*a_k x^(k)) + (∑(k=0) -2a_k x^k) = 0

Next, re-index the sums to each have x to the same exponent.

Substitute n=k-2 for the first sum

∑(k=2) (k-1)*k*a_k x^(k-2) = ∑(n=0) (n+2-1)*(n+2)*a_(n+2) x^(n+2-2)

= ∑(n=0) (n+1)*(n+2)*a_(n+2) x^(n)

Switching back to k and re-substituting gives

∑(k=0) (k+1)*(k+2)*a_(k+2) x^(k) + (∑(k=1) -2*k*a_k x^(k)) + (∑(k=0) -2a_k x^k) = 0

We next want all of the sums to start at the same index, so evaluate the 0'th terms of the first and third sums

1*2a_2*x^0 -2a_0*x^0 +∑(k=1) (k+1)*(k+2)*a_(k+2) x^(k) + (∑(k=1) -2*k*a_k x^(k)) + (∑(k=1) -2a_k x^k) = 0

2(a_2-a_0) + ∑(k=1) (k+1)*(k+2)*a_(k+2) x^(k) + (∑(k=1) -2*k*a_k x^(k) + (∑(k=1) -2a_k x^k) = 0

Finally, combine the sums together.

2(a_2-a_0) + ∑(k=1) (k+1)*(k+2)*a_(k+2) x^(k) -2*k*a_k x^(k) -2a_k x^k = 0

2(a_2-a_0) + ∑(k=1) [(k+1)*(k+2)*a_(k+2) - 2*k*a_k - 2a_k] x^k = 0

We can equate the coefficients of x^k on both sides of the equation.

Constant term: 2(a_2-a_0)=0 => a_0 = a_2

Rest: (k+1)*(k+2)*a_(k+2) - 2*k*a_k - 2a_k = 0 = (k+1)(k+2)*a_(k+2) - 2*(k+1)*a_k = 0

=> a_(k+2) = a_k*(2)/(k+2)

Let a_0 = H and a_1 = K be the arbitrary constants

Testing terms this gives

a_2 = H

a_3 = 2a_1/3 = 2K/3

a_4 = 2a_2/4 = H/2

a_5 = 2a_3/5 = 2^2*K/(3*5)

a_6 = 2a_4/6 = H/(2*3)

a_7 = 2a_5/7 = 2^3*K/(3*5*7)

a_8 = 2a_6/8 = H/(2*3*4)

a_9 = 2a_7/9 = 2^4*K/(3*5*7*9)

a_10 = 2a_8/10 = H=(2*3*4*5)

a_11 = 2a_9/11 = 2^5*K/(3*5*7*9*11)

From here, you can see a pattern for the even and odd terms:

a_{2n} = H/(n!)

a_{2n+1} = K(2^n)/((2n+1)!!) (The !! is the "double factorial" notation, where n!! = n(n-2)(n-4)...5*3*1)

Going back to the original for of the homogeneous solution:

y=∑(k=0) a_k x^k

The sum can be split into even and odd portions:

y_h = H*(1 + ∑(k=1) [(x^{2n})/(n!)]) + K*(x + ∑(k=1) [(2^n*x^{2n+1})/((2n+1)!!)])

These are two linearly independent solutions, so they represent all solutions to the homogeneous equation.

Now, combining the homogeneous and particular solutions gives

y = y_P + y_h

y = (-x/4) + H*(1 + ∑(k=1) [(x^{2n})/(n!)]) + K*(x + ∑(k=1) [(2^n*x^{2n+1})/((2n+1)!!)])

This is the solution to the differential equation in power series form.

Answer 2

This is a linear differential equation of second order with the corresponding homogeneous

being y''-2xy'-2y= 0

Let us solve the homogeneous one using the power series method ,

Let y = ∑_n=0 a_n xⁿ . Then y' = ∑_n=1 n a_n x^n-1 and y'' = ∑_n=2 n(n-1) a_n x^n-2

and going through the hoops and the loops you will and up with y = 1 + x² + x⁴/ 2! + x⁶/3! + x⁸/4 +...

that is y = e^{x^2}.

Given that y = e^{x^2}. is a solution to y''-2xy'-2y= 0 it is doable to find a second solution for the y''-2xy'-2y= 0

by proposing a solution of the form y = u(x) e^{x^2}.