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

cos space open parentheses dy over dx close parentheses space space equals straight a space space space left parenthesis straight a space element of space straight R right parenthesis semicolon space space space straight y space equals space 1 space space when space straight x space equals space 0

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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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The given differential equation is

straight x squared dy over dx space equals space straight x squared plus 5 xy plus 4 straight y squared

dy over dx space equals space fraction numerator straight x squared plus 5 xy plus 4 straight y squared over denominator straight x squared end fraction

Put y = v x so that

dy over dx equals straight v plus straight x dv over dx

therefore space space space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx space equals fraction numerator straight x squared plus 5 straight x. space straight v space straight x space plus space 4 space straight v squared space straight x squared over denominator straight x squared end fraction

straight v plus straight x dv over dx space equals space 1 plus 5 straight v plus 4 straight v squared

straight x dv over dx space equals space 4 space straight v squared plus 4 space straight v space plus space 1

Separating the variables, we get,

fraction numerator 1 over denominator 4 space straight v squared plus 4 space straight v space plus space 1 end fraction dv space equals space 1 over straight x dx

Integrating

integral fraction numerator 1 over denominator 4 straight v squared plus 4 straight v plus 1 end fraction straight d space straight v space equals space integral 1 over straight x dx

integral fraction numerator 1 over denominator left parenthesis 2 space straight v space plus space 1 right parenthesis squared end fraction dv space equals space integral 1 over straight x dx space space space space space or space space space space integral left parenthesis 2 straight v plus 1 right parenthesis to the power of negative 2 end exponent space dv space equals space integral 1 over straight x dx

therefore space space space space fraction numerator left parenthesis 2 space straight v space plus space 1 right parenthesis to the power of negative 1 end exponent over denominator left parenthesis 2 right parenthesis space left parenthesis negative 1 right parenthesis end fraction space equals space log space open vertical bar straight x close vertical bar space plus straight c space space space space space or space space space space minus fraction numerator 1 over denominator 2 space left parenthesis 2 space straight v space plus 1 right parenthesis end fraction log space open vertical bar straight x close vertical bar plus straight c

negative fraction numerator 1 over denominator 2 open parentheses 2 begin display style straight y over straight x end style plus 1 close parentheses end fraction space equals space log space open vertical bar straight x close vertical bar plus straight c space space space space space space space space or space space space space space space space space space minus fraction numerator straight x over denominator 2 space left parenthesis straight x plus 2 space straight y right parenthesis end fraction space equals space log space open vertical bar straight x close vertical bar plus straight c

fraction numerator straight x over denominator 2 space left parenthesis straight x plus 2 space straight y right parenthesis end fraction plus log space open vertical bar straight x close vertical bar plus straight c space equals space 0

is the required solution.

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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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