Question

(x^2 + 1)y''(x) + xy'(x) + xy(x) = 0, y(0) = 3, y'(0) = -1

Answer 1

ANSWER on Question #82285 – Math – Differential Equations

QUESTION

Find a power series solution of the initial-value problem.

$$(x^2 + 1) y''(x) + x y'(x) + x y(x) = 0, \quad y(0) = 3, \quad y'(0) = -1$$

SOLUTION

Let

$$y(x) = \sum_{n=0}^{+\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$$ $$y(0) = 3 = a_0 + a_1 \cdot 0 + a_2 \cdot 0^2 + a_3 \cdot 0^3 + \cdots \rightarrow \boxed{a_0 = 3}$$ $$y'(0) = -1 = a_1 + 2 \cdot a_2 \cdot 0 + 3 \cdot a_3 \cdot 0^2 + \cdots \rightarrow \boxed{a_1 = -1}$$

Then,

$$y(x) = \sum_{n=0}^{+\infty} a_n x^n \rightarrow y'(x) = \sum_{n=0}^{+\infty} a_n \cdot n \cdot x^{n-1} \equiv \sum_{n=1}^{+\infty} a_n \cdot n \cdot x^{n-1} = \begin{bmatrix} n-1 = k \\ n = k+1 \\ n=1 \rightarrow k=0 \end{bmatrix} \rightarrow$$ $$y'(x) = \sum_{k=0}^{+\infty} a_{(k+1)} \cdot (k+1) \cdot x^k \rightarrow \boxed{y'(x) = \sum_{n=0}^{+\infty} a_{(n+1)} \cdot (n+1) \cdot x^n}$$ $$y'(x) = \sum_{n=0}^{+\infty} a_{(n+1)} \cdot (n+1) \cdot x^n \rightarrow y''(x) = \sum_{n=0}^{+\infty} a_{(n+1)} \cdot (n+1) \cdot n \cdot x^{n-1} \equiv$$ $$\equiv \sum_{n=1}^{+\infty} a_{(n+1)} \cdot (n+1) \cdot n \cdot x^{n-1} = \begin{bmatrix} n-1 = k \\ n = k+1 \\ n=1 \rightarrow k=0 \end{bmatrix} = \sum_{k=0}^{+\infty} a_{(k+2)} \cdot (k+2)(k+1) \cdot x^k \rightarrow$$ $$y''(x) = \sum_{k=0}^{+\infty} a_{(k+2)} \cdot (k+2)(k+1) \cdot x^k \rightarrow \boxed{y''(x) = \sum_{n=0}^{+\infty} a_{(n+2)} \cdot (n+2)(n+1) \cdot x^n}$$

Substitute all the found series in the original equation.

$$(x ^ {2} + 1) y ^ {\prime \prime} (x) + x y ^ {\prime} (x) + x y (x) = 0 \rightarrow$$ $$\begin{array}{l} (x ^ {2} + 1) \cdot \sum_ {n = 0} ^ {+ \infty} a _ {(n + 2)} \cdot (n + 2) (n + 1) \cdot x ^ {n} + x \cdot \sum_ {n = 0} ^ {+ \infty} a _ {(n + 1)} \cdot (n + 1) \cdot x ^ {n} + x \cdot \sum_ {n = 0} ^ {+ \infty} a _ {n} x ^ {n} = 0 \rightarrow \\ \sum_ {n = 0} ^ {+ \infty} a _ {(n + 2)} \cdot (n + 2) (n + 1) \cdot x ^ {n + 2} + \sum_ {n = 0} ^ {+ \infty} a _ {(n + 2)} \cdot (n + 2) (n + 1) \cdot x ^ {n} + \\ + \sum_ {n = 0} ^ {+ \infty} a _ {(n + 1)} \cdot (n + 1) \cdot x ^ {n + 1} + \sum_ {n = 0} ^ {+ \infty} a _ {n} x ^ {n + 1} = 0 \\ \end{array}$$

In these series are different degrees of $x$ . Let us make sure that all ranks have the same degrees.

$$\sum_ {n = 0} ^ {+ \infty} a _ {(n + 2)} \cdot (n + 2) (n + 1) \cdot x ^ {n + 2} = \left[ \begin{array}{c} n + 2 = k \\ n = k - 2 \\ n = 2 \to k = 2 \end{array} \right] = \sum_ {k = 2} ^ {+ \infty} a _ {k} \cdot k (k - 1) \cdot x ^ {k} \to$$ $$\sum_ {n = 0} ^ {+ \infty} a _ {(n + 2)} \cdot (n + 2) (n + 1) \cdot x ^ {n + 2} = \sum_ {k = 2} ^ {+ \infty} a _ {k} \cdot k (k - 1) \cdot x ^ {k}$$ $$\begin{array}{l} \sum_ {n = 0} ^ {+ \infty} a _ {(n + 2)} \cdot (n + 2) (n + 1) \cdot x ^ {n} = a _ {(0 + 2)} \cdot (0 + 2) (0 + 1) \cdot x ^ {0} + a _ {(1 + 2)} \cdot (1 + 2) (1 + 1) \cdot x ^ {1} + \\ + \sum_ {k = 2} ^ {+ \infty} a _ {(k + 2)} \cdot (k + 2) (k + 1) \cdot x ^ {k} = 2 a _ {2} + 6 a _ {3} x + \sum_ {k = 2} ^ {+ \infty} a _ {(k + 2)} \cdot (k + 2) (k + 1) \cdot x ^ {k} \rightarrow \\ \end{array}$$ $$\sum_ {n = 0} ^ {+ \infty} a _ {(n + 2)} \cdot (n + 2) (n + 1) \cdot x ^ {n} = 2 a _ {2} + 6 a _ {3} x + \sum_ {k = 2} ^ {+ \infty} a _ {(k + 2)} \cdot (k + 2) (k + 1) \cdot x ^ {k}$$ $$\sum_ {n = 0} ^ {+ \infty} a _ {(n + 1)} \cdot (n + 1) \cdot x ^ {n + 1} = \left[ \begin{array}{c} n + 1 = m \\ n = m - 1 \\ n = 0 \to m = 1 \end{array} \right] = \sum_ {m = 1} ^ {+ \infty} a _ {m} \cdot m \cdot x ^ {m} = a _ {1} x + \sum_ {m = 2} ^ {+ \infty} a _ {m} \cdot m \cdot x ^ {m} \to$$ $$\sum_ {n = 0} ^ {+ \infty} a _ {(n + 1)} \cdot (n + 1) \cdot x ^ {n + 1} = a _ {1} x + \sum_ {k = 2} ^ {+ \infty} a _ {k} \cdot k \cdot x ^ {k}$$ $$\sum_ {n = 0} ^ {+ \infty} a _ {n} x ^ {n + 1} = \left[ \begin{array}{c} n + 1 = m \\ n = m - 1 \\ n = 0 \to m = 1 \end{array} \right] = \sum_ {m = 1} ^ {+ \infty} a _ {(m - 1)} x ^ {m} = a _ {(1 - 1)} x + \sum_ {m = 2} ^ {+ \infty} a _ {(m - 1)} x ^ {m} \to$$ $$\sum_{n=0}^{+\infty} a_n x^{n+1} = a_0 x + \sum_{k=2}^{+\infty} a_{(k-1)} x^k$$

Our equation takes the form

$$\sum_{k=2}^{+\infty} a_k \cdot k(k-1) \cdot x^k + 2a_2 + 6a_3 x + \sum_{k=2}^{+\infty} a_{(k+2)} \cdot (k+2)(k+1) \cdot x^k +$$ $$+ a_1 x + \sum_{k=2}^{+\infty} a_k \cdot k \cdot x^k + a_0 x + \sum_{k=2}^{+\infty} a_{(k-1)} x^k = 0 \rightarrow$$ $$2a_2 + (6a_3 + a_1 + a_0) x + \sum_{k=2}^{+\infty} [k(k-1)a_k + k a_k + a_{(k-1)} + (k+2)(k+1)a_{(k+2)}] x^k = 0 \rightarrow$$ $$\left\{ \begin{array}{c} a_0 = 3 \\ a_1 = -1 \\ a_2 = 0 \\ 6a_3 + 3 - 1 = 0 \rightarrow a_3 = -\frac{1}{3} \\ a_{(k+2)} = -\frac{k^2 a_k + a_{(k-1)}}{(k+1)(k+2)}, \quad \forall k \geq 2 \end{array} \right.$$

Conclusion,

$$y(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \sum_{k=4}^{+\infty} a_k x^k = 3 - x - \frac{x^3}{3} + \sum_{k=4}^{+\infty} a_k x^k$$ $$\left\{ \begin{array}{c} a_0 = 3 \\ a_1 = -1 \\ a_2 = 0 \\ a_3 = -\frac{1}{3} \\ a_{(k+2)} = -\frac{k^2 a_k + a_{(k-1)}}{(k+1)(k+2)}, \quad \forall k \geq 2 \\ y(x) = 3 - x - \frac{x^3}{3} + \sum_{k=4}^{+\infty} a_k x^k \end{array} \right.$$

ANSWER

$$\left\{ \begin{array}{c} a _ {0} = 3 \\ a _ {1} = - 1 \\ a _ {2} = 0 \\ a _ {3} = - \frac {1}{3} \\ a _ {(k + 2)} = - \frac {k ^ {2} a _ {k} + a _ {(k - 1)}}{(k + 1) (k + 2)}, \quad \forall k \geq 2 \\ y (x) = 3 - x - \frac {x ^ {3}}{3} + \sum_ {k = 4} ^ {+ \infty} a _ {k} x ^ {k} \end{array} \right.$$

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