Use the guide to solve for the taylor expansion series. dy= f(x,y) = 3x – 2y with y(0) = 0dxAnalytical Solution:dy+ 2y = 3xdxv= e2xe2xy = 3| xe2xdx + CLet u = x; du = dx= | e2*dx = ;e2xetty = xe* - e*dx + CV =33e 2x dx + C2.333.e2*y =xe2x -e2* + C ==(2xe²x – e2x) + Cy =7(2x – 1) + Ce-2*; y(0) = 030 = (2.0 – 1) + Ce°;C = =33(2x – 1) +e-:4y = 7Using TSE:dy= f(x, y) = 3x – 2ydxf = 3x – 2yf' = ax" ay'f = 3 + (-2)(3x – 2y) = 3 – 6x + 4y+f" =of +Lf = -6+4(3x – 2y) = -6+12x – 8yaxдуaf" áf"f" :-f = 12 + (-8)(3x – 2y) = 12 – 24x + 16y+axдуh3h2Yi+ı = yi + h(3x; – 2y;) +(3 – 6x; + 4y;) +(-6+ 12x; – 8y;)h*-(12 — 24х; + 16у)24 dy+y = e2xe 2xdx1.2 ye2*dx = 4 + e2*dy2.

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Answered: dy dx +y = e2x 1.

Math

Calculus

dy dx +y = e2x 1.

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter6: Applications Of The Derivative

Section6.CR: Chapter 6 Review

Problem 25CR

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Question

Use the guide to solve for the taylor expansion series.

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
+y = e2x
e 2x
dx
1.
2 ye2*dx = 4 + e2*dy
2.

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