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Math  /  Calculus

QuestionUse the power series to solve the following Differential Equation up to its indicial equation xd2ydx2+dydx+x2y=0x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x^{2} y=0 (6 marks)

Studdy Solution

STEP 1

1. We are solving the differential equation using power series methods.
2. The solution y(x) y(x) can be expressed as a power series centered at x=0 x = 0 .
3. The differential equation is linear and has variable coefficients.

STEP 2

1. Assume a power series solution for y(x) y(x) .
2. Differentiate the power series to find expressions for dydx\frac{d y}{d x} and d2ydx2\frac{d^2 y}{d x^2}.
3. Substitute the power series expressions into the differential equation.
4. Collect terms and find the recurrence relation.
5. Derive the indicial equation from the lowest power of x x .

STEP 3

Assume a power series solution for y(x) y(x) :
y(x)=∑n=0∞anxn y(x) = \sum_{n=0}^{\infty} a_n x^n
where an a_n are coefficients to be determined.

STEP 4

Differentiate the power series to find expressions for dydx\frac{d y}{d x} and d2ydx2\frac{d^2 y}{d x^2}:
dydx=∑n=1∞nanxn−1 \frac{d y}{d x} = \sum_{n=1}^{\infty} n a_n x^{n-1}
d2ydx2=∑n=2∞n(n−1)anxn−2 \frac{d^2 y}{d x^2} = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}

STEP 5

Substitute the power series expressions into the differential equation:
x∑n=2∞n(n−1)anxn−2+∑n=1∞nanxn−1+x2∑n=0∞anxn=0 x \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + \sum_{n=1}^{\infty} n a_n x^{n-1} + x^2 \sum_{n=0}^{\infty} a_n x^n = 0
Simplify each term:
1. x∑n=2∞n(n−1)anxn−2=∑n=2∞n(n−1)anxn−1 x \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-1}
2. ∑n=1∞nanxn−1 \sum_{n=1}^{\infty} n a_n x^{n-1} remains unchanged.
3. x2∑n=0∞anxn=∑n=0∞anxn+2 x^2 \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{n+2}

STEP 6

Align the indices of the series to combine them:
1. For the first term, ∑n=2∞n(n−1)anxn−1 \sum_{n=2}^{\infty} n(n-1) a_n x^{n-1} , let n−1=m n-1 = m so n=m+1 n = m+1 , then:
∑m=1∞(m+1)mam+1xm \sum_{m=1}^{\infty} (m+1)m a_{m+1} x^m
2. For the second term, ∑n=1∞nanxn−1 \sum_{n=1}^{\infty} n a_n x^{n-1} , let n−1=m n-1 = m so n=m+1 n = m+1 , then:
∑m=0∞(m+1)am+1xm \sum_{m=0}^{\infty} (m+1) a_{m+1} x^m
3. For the third term, ∑n=0∞anxn+2 \sum_{n=0}^{\infty} a_n x^{n+2} , let n+2=m n+2 = m so n=m−2 n = m-2 , then:
∑m=2∞am−2xm \sum_{m=2}^{\infty} a_{m-2} x^m
Combine the series:
∑m=1∞(m+1)mam+1xm+∑m=0∞(m+1)am+1xm+∑m=2∞am−2xm=0 \sum_{m=1}^{\infty} (m+1)m a_{m+1} x^m + \sum_{m=0}^{\infty} (m+1) a_{m+1} x^m + \sum_{m=2}^{\infty} a_{m-2} x^m = 0

STEP 7

Collect terms and find the recurrence relation by setting the coefficient of each power of x x to zero:
For m=0 m = 0 :
(1)a1=0 (1) a_1 = 0
For m≥1 m \geq 1 :
(m+1)mam+1+(m+1)am+1+am−2=0 (m+1)m a_{m+1} + (m+1) a_{m+1} + a_{m-2} = 0
Simplify to find the recurrence relation:
(m+1)(m+1)am+1+am−2=0 (m+1)(m+1) a_{m+1} + a_{m-2} = 0

STEP 8

Derive the indicial equation from the lowest power of x x :
The indicial equation is derived from the lowest power of x x , which is when m=0 m = 0 :
a1=0 a_1 = 0
This implies that the indicial equation is trivial in this case, as the lowest power does not provide a non-zero condition.
The indicial equation is trivial, indicating no additional conditions from the lowest power of x x .

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