238. [T] sin(x)y + cos(x) = 2x 239. [T] √x² + 1y = y +2 240. [T] xy+2r2y=x+1 Solve the following initial-value problems by using integrating factors. 241. y' + y = x, y(0) = 3 242. y' = y + 2x², y(0) = 0 243. xy' = y - 3x³, y(1) = 0 244. x²y = xy-lnx, y(1) = 1 245. (1+x²)y=y-1, y(0) = 0 246. xy = y + 2xlnx, y(1) = 5 wing B
238. [T] sin(x)y + cos(x) = 2x 239. [T] √x² + 1y = y +2 240. [T] xy+2r2y=x+1 Solve the following initial-value problems by using integrating factors. 241. y' + y = x, y(0) = 3 242. y' = y + 2x², y(0) = 0 243. xy' = y - 3x³, y(1) = 0 244. x²y = xy-lnx, y(1) = 1 245. (1+x²)y=y-1, y(0) = 0 246. xy = y + 2xlnx, y(1) = 5 wing B
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
Section6.3: Implicit Differentiation
Problem 11E
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Solve 241 ,243
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238. [T] sin(x)y' + cos(x)y = 2x
239. [T] x² + ly = y +2
240. [T] x*y+2r2y=x+1
Solve the following initial-value problems by using
integrating factors.
241. y' + y = x, y(0) = 3
242. y' = y + 2x², y(0) = 0
243. xy = y - 3x³, y(1) = 0
244. x²y = xy-lnx, y(1) = 1
245. (1+x²)y=y-1, y(0) = 0
246. xy' = y + 2xlnx, y(1) = 5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03fd5f25-eded-4046-9c65-9e89b5e28b89%2Fe892fed2-da4a-4c2a-87c1-bc9d7b410ff7%2Fyozt0c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:owing
238. [T] sin(x)y' + cos(x)y = 2x
239. [T] x² + ly = y +2
240. [T] x*y+2r2y=x+1
Solve the following initial-value problems by using
integrating factors.
241. y' + y = x, y(0) = 3
242. y' = y + 2x², y(0) = 0
243. xy = y - 3x³, y(1) = 0
244. x²y = xy-lnx, y(1) = 1
245. (1+x²)y=y-1, y(0) = 0
246. xy' = y + 2xlnx, y(1) = 5
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