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181.

Solve:
$cos space left parenthesis straight x plus straight y right parenthesis space dy over dx space equals space 1.$

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182.

Solve:
$dy over dx space equals space tan space left parenthesis straight x plus straight y right parenthesis$

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183. Solve the following initial value problem:
(x + y + 1)² dy = dx, y ( –1) = 0

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184.

Solve the following initial value problem
cos (x + y) dy = dx, y (0) = 0

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185. For the following differential equation, given below, find particular solution satisfying the given condition:

open parentheses straight x cubed plus straight x squared plus straight x plus 1 close parentheses dy over dx space equals space 2 straight x squared plus straight x semicolon space space straight y space equals space 1 space space when space straight x space equals space 0

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186. For the following differential equation, given below, find particular solution satisfying the given condition:

straight x open parentheses straight x squared minus 1 close parentheses space dy over dx space equals space 1 space semicolon space space straight y space equals space 0 space space when space straight x space equals space 2

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The given differential equation is

cos space open parentheses dy over dx close parentheses space equals space straight a space space space space or space space space dy over dx space equals space cos to the power of negative 1 end exponent straight a

therefore space space space space space space space space space dy space equals space cos to the power of negative 1 end exponent straight a. space dx space space space rightwards double arrow space space space space integral space dy space equals space cos to the power of negative 1 end exponent space straight a integral space dx

therefore space space space space space space space space straight y space equals space left parenthesis cos to the power of negative 1 end exponent straight a right parenthesis. space straight x plus space straight c

...(1)
Now y = 1 when x = 0

therefore space space space 1 space equals space left parenthesis cos to the power of negative 1 end exponent straight a right parenthesis space. space 0 space plus straight c space rightwards double arrow space space 1 space equals space straight c therefore space space space space from space left parenthesis 1 right parenthesis comma space space straight y space equals space straight x space cos to the power of negative 1 end exponent straight a plus 1

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188. For the following differential equation, given below, find particular solution satisfying the given condition:

dy over dx equals straight y space tanx space semicolon space space straight y space equals space 1 space space when space straight x space equals space 0

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189.

Solve: $straight x squared dy over dx space equals straight x squared plus 5 xy plus 4 straight y squared.$

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190.

Solve the following differential equation:
$straight x squared dy over dx space equals space 2 xy plus straight y squared$

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