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dy 3D f (x, у) — y sес х with y(0) — 2 dx

dy 3D f (x, у) — y sес х with y(0) — 2 dx

Advanced Engineering Mathematics
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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1. I have already solved for the analytical solution. Solve for the taylor expansion series. Use the guide. 

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dy = f(x,y) = 3x – 2y with y(0) = 0 dx Analytical Solution: dy + 2y = 3x dx v= e2x e2xy = 3| xe2xdx + C Let u = x; du = dx = | e2*dx = ;e2x etty = xe* - e*dx + C V = 3 3 e 2x dx + C 2. 3 3 3. e2*y =xe2x -e2* + C ==(2xe²x – e2x) + C y =7(2x – 1) + Ce-2*; y(0) = 0 3 0 = (2.0 – 1) + Ce°;C = = 3 3 (2x – 1) +e-: 4 y = 7 Using TSE: dy = f(x, y) = 3x – 2y dx f = 3x – 2y f' = ax" ay' f = 3 + (-2)(3x – 2y) = 3 – 6x + 4y + f" = of +Lf = -6+4(3x – 2y) = -6+12x – 8y ax ду af" áf" f" : -f = 12 + (-8)(3x – 2y) = 12 – 24x + 16y + ax ду h3 h2 Yi+ı = yi + h(3x; – 2y;) +(3 – 6x; + 4y;) +(-6+ 12x; – 8y;) h* -(12 — 24х; + 16у) 24
Transcribed Image Text:dy = f(x,y) = 3x – 2y with y(0) = 0 dx Analytical Solution: dy + 2y = 3x dx v= e2x e2xy = 3| xe2xdx + C Let u = x; du = dx = | e2*dx = ;e2x etty = xe* - e*dx + C V = 3 3 e 2x dx + C 2. 3 3 3. e2*y =xe2x -e2* + C ==(2xe²x – e2x) + C y =7(2x – 1) + Ce-2*; y(0) = 0 3 0 = (2.0 – 1) + Ce°;C = = 3 3 (2x – 1) +e-: 4 y = 7 Using TSE: dy = f(x, y) = 3x – 2y dx f = 3x – 2y f' = ax" ay' f = 3 + (-2)(3x – 2y) = 3 – 6x + 4y + f" = of +Lf = -6+4(3x – 2y) = -6+12x – 8y ax ду af" áf" f" : -f = 12 + (-8)(3x – 2y) = 12 – 24x + 16y + ax ду h3 h2 Yi+ı = yi + h(3x; – 2y;) +(3 – 6x; + 4y;) +(-6+ 12x; – 8y;) h* -(12 — 24х; + 16у) 24
dy = f(x, y) = y sec x with y(0) = 2 dx Analytical solution dy = y sec x dx dy sec x dx y In y = In(sec x + tan x) + C eln y = eln(secx+tan x)+C eln y = eln(secx+tan x)eC y = C(sec x + tan x) when x = 0, y = 2 2 = C(sec 0 + tan 0) 2 = C(1+ 0) C = 2 General Solution: y = 2(sec x + tan x)
Transcribed Image Text:dy = f(x, y) = y sec x with y(0) = 2 dx Analytical solution dy = y sec x dx dy sec x dx y In y = In(sec x + tan x) + C eln y = eln(secx+tan x)+C eln y = eln(secx+tan x)eC y = C(sec x + tan x) when x = 0, y = 2 2 = C(sec 0 + tan 0) 2 = C(1+ 0) C = 2 General Solution: y = 2(sec x + tan x)
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