answer both questions Q3. Use the power series to solve the following ODE:22y+ 3rdr?dy+(-1+2)y 0Q4. Solve the following PDEUyy –Uy – 12u = 2e3y

Name: answer both questions Q3. Use the power series to solve the following
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Description: answer both questions Q3. Use the power series to solve the following ODE:22y+ 3rdr?dy+(-1+2)y 0Q4. Solve the following PDEUyy –Uy – 12u = 2e3y

Given: 3) 2x2d2ydx2+3xdydx+(-1+x)y=04) uyy-uy-12u=2e3y

Answered: 3. Use the power series to solve the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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answer both questions

Q3. Use the power series to solve the following ODE:
22y
+ 3r
dr?
dy
+(-1+2)y 0
Q4. Solve the following PDE
Uyy –
Uy – 12u = 2e3y

Expert Solution

Step 1

Given:

$3) 2 x^{2} \frac{d^{2} y}{d x^{2}} + 3 x \frac{d y}{d x} + (- 1 + x) y = 0 4) u_{y y} - u_{y} - 12 u = 2 e^{3 y}$

Step 2

$3) 2 x^{2} \frac{d^{2} y}{d x^{2}} + 3 x \frac{d y}{d x} + (- 1 + x) y = 0 . . . . . . (1)$

Consider,

ordinary point of the equation x=0

$T h e n, y = \sum_{n = 0}^{\infty} a n x^{n} T h e r e f o r e, \frac{d y}{d x} = {\sum n}_{n = 1}^{\infty} a n x^{n - 1}, \frac{d^{2} y}{d x^{2}} = {\sum n (n - 1)}_{n = 2}^{\infty} a n x^{n - 2}$

Substituting the values in equation (1)

$2 \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} + 3 \sum_{n = 1}^{\infty} n a_{n} x^{n} - \sum_{n = 0}^{\infty} a_{n} x^{n} - \sum_{n}^{\infty} a_{n} x^{n + 1} = 0$

$2 n (n - 1) a_{n} + 3 n a_{n} - a_{n - 1} = 0 T h e r e f o r e, [\sum_{n = 0}^{\infty} a_{n} x^{n + 1} = \sum_{n = 1}^{\infty} a_{n - 1} x^{n}] = (2 n^{2} - 2 n + 3 n) a_{n} = a_{n - 1} \Rightarrow a_{n} = \frac{a_{n - 1}}{(2 n^{2} + n)} T h e r e f o r e, a_{1} = \frac{a_{0}}{3}, a_{2} = \frac{a_{1}}{10} s o, y = (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4}) B y s u b s t i t u t i n g t h e v a l u e s, = (a_{0} + \frac{a_{0}}{3} x + \frac{a_{1}}{10} x^{2} + \frac{a_{2}}{21} x^{3} + \frac{a_{3}}{36} x^{4} + . . . .) = \frac{4 a_{0}}{3} x + \frac{a_{1}}{10} x^{2} + \frac{a_{2}}{21} x^{3} + \frac{a_{3}}{36} x^{4} + . . . .$

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