20. Verify directly thatr2 dy dxdx dy1+x - yJo J2 1+x – y

take Left Hand side ∫23∫02dydx1+x-y the integral can be written as ∫23∫02dydx1+x-y=∫23∫02dy1+x-ydx…

Answered: 20. Verify directly that r2 dy dx dx dy…

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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Related questions

Question

20. Verify directly that
r2 dy dx
dx dy
1+x - y
Jo J2 1+x – y

Expert Solution

Step 1

take Left Hand side $\int_{2}^{3} \int_{0}^{2} \frac{d y d x}{1 + x - y}$

the integral can be written as $\int_{2}^{3} \int_{0}^{2} \frac{d y d x}{1 + x - y} = \int_{2}^{3} (\int_{0}^{2} \frac{d y}{1 + x - y}) d x$

first solve the inner integral

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

substitute $u = 1 + x - y$

$d u = - d y$

when

$y = 0, u = 1 + x y = 2, u = 1 + x - 2$

therefore,

$\begin{array}{rcl} \int_{0}^{2} \frac{d y}{1 + x - y} & = & - \int_{1 + x}^{- 1 + x} \frac{d u}{u} \\ = & - {(\ln |u|)}_{1 + x}^{- 1 + x} \\ = & - \{\ln |- 1 + x| - \ln |1 + x|\} \\ = & - \ln |- 1 + x| + \ln |1 + x| \\ = & - \ln |x - 1| + \ln |x + 1| \end{array}$

now

$\begin{array}{rcl} \int_{2}^{3} \int_{0}^{2} \frac{d y d x}{1 + x - y} & = & \int_{2}^{3} (\int_{0}^{2} \frac{d y}{1 + x - y}) d x \\ = & \int_{2}^{3} - (\ln |x - 1| + \ln |x + 1|) d x \\ = & - \int_{2}^{3} (\ln |x - 1|) d x + \int_{2}^{3} \ln |x + 1| d x \end{array}$

solve the integral $\int_{2}^{3} (\ln |x - 1|) d x + \int_{2}^{3} \ln |x + 1| d x$

Step 2

substitute

$\begin{array}{rcl} u & = & x - 1 \\ d u & = & d x \end{array}$

when

$x = 2, u = 1 x = 3, u = 2$

therefore, the integral becomes

$\begin{array}{rcl} \int_{2}^{3} (\ln |x - 1|) d x & = & \int_{1}^{2} \ln u d u \end{array}$

integrate by parts by taking parts as $\ln u a n d 1$

$\begin{array}{rcl} \int_{2}^{3} (\ln |x - 1|) d x & = & \int_{1}^{2} \ln u d u \\ = & {(\ln u (u))}_{1}^{2} - \int_{1}^{2} u \frac{1}{u} d u \\ = & {(\ln u (u))}_{1}^{2} - \int_{1}^{2} 1 d u \\ = & {(\ln u (u))}_{1}^{2} - {(u)}_{1}^{2} \\ = & (\ln 2 (2) - \ln (1) 1) - (2 - 1) \\ = & (2 \ln 2 - 0) - (1) \\ = & 2 \ln 2 - 1 \end{array}$

substitute

$\begin{array}{rcl} u & = & x + 1 \\ d u & = & d x \end{array}$

when

$x = 2, u = 3 x = 3, u = 4$

therefore, the integral becomes

$\begin{array}{rcl} \int_{2}^{3} (\ln |x + 1|) d x & = & \int_{3}^{4} \ln u d u \end{array}$

integrate by parts by taking parts as $\ln u a n d 1$

$\begin{array}{rcl} \int_{2}^{3} (\ln |x + 1|) d x & = & \int_{3}^{4} \ln u d u \\ = & {(\ln u (u))}_{3}^{4} - \int_{3}^{4} u \frac{1}{u} d u \\ = & {(\ln u (u))}_{3}^{4} - \int_{3}^{4} 1 d u \\ = & {(\ln u (u))}_{3}^{4} - {(u)}_{3}^{4} \\ = & (\ln 4 (4) - \ln (3) 3) - (4 - 3) \\ = & (4 \ln 4 - 3 \ln 3) - (1) \\ = & (4 \ln 2^{2} - 3 \ln 3) - (1) \\ = & (8 l n 2 - 3 \ln 3) - (1) \end{array}$

Step 3

therefore,

$\begin{array}{rcl} \int_{2}^{3} \int_{0}^{2} \frac{d y d x}{1 + x - y} & = & - \int_{2}^{3} (\ln |x - 1|) d x + \int_{2}^{3} \ln |x + 1| d x \\ = & - (2 \ln 2 - 1) + 8 \ln 2 - 3 \ln 3 - 1 \\ = & - 2 \ln 2 + 1 + 8 \ln 2 - 3 \ln 3 - 1 \\ = & 6 \ln 2 - 3 \ln 3 \end{array}$