Find the solutions for the following differential equations and initial value problems. If a problem can be solved both by methods for linear equations and for separable equations, solve it both ways and compare the answers. (a) dy dt t³ - 2y t dy (b) t +ty=1-y, dt dy (c) t + 2y dt (d) +y dt sint 1+ et y(1) = 0 y (2) = 1
Find the solutions for the following differential equations and initial value problems. If a problem can be solved both by methods for linear equations and for separable equations, solve it both ways and compare the answers. (a) dy dt t³ - 2y t dy (b) t +ty=1-y, dt dy (c) t + 2y dt (d) +y dt sint 1+ et y(1) = 0 y (2) = 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.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 12CR
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Needed to be solved Part A,B,C and D correctly in 30 minutes and get the thumbs up please show neat and clean work for it please don't copy from Chegg please
![! Find the solutions for the following differential equations and initial value problems. If a
problem can be solved both by methods for linear equations and for separable equations, solve it both
ways and compare the answers.
(a)
dy
dt
(b) t +ty=1-y,
dt
dy
dt
(c) t
(d) du
dy
(e)
(f)
dy
dt
t³ - 2y
t
+2y= =
+y=
-
sin t
1+ et
dy
1²-1
dt y² +1'
(g) y+t=0,
+3y
y(1) = 0
y (2) = 1
y(-1) = 1
t> 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e51f0f4-140a-4430-91a7-57800eabe18a%2F4386c90a-5634-4cb5-9ac5-51a4982d9653%2F7icbpc6j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:! Find the solutions for the following differential equations and initial value problems. If a
problem can be solved both by methods for linear equations and for separable equations, solve it both
ways and compare the answers.
(a)
dy
dt
(b) t +ty=1-y,
dt
dy
dt
(c) t
(d) du
dy
(e)
(f)
dy
dt
t³ - 2y
t
+2y= =
+y=
-
sin t
1+ et
dy
1²-1
dt y² +1'
(g) y+t=0,
+3y
y(1) = 0
y (2) = 1
y(-1) = 1
t> 0
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