For the following differential equation, given below, find  par

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 Multiple Choice QuestionsShort Answer Type

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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 Multiple Choice QuestionsLong Answer Type

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


The given differential equation is
                       open parentheses straight x cubed plus straight x squared plus straight x plus 1 close parentheses space dy over dx space equals space 2 straight x squared plus straight x
or             dy over dx space equals space fraction numerator 2 straight x squared plus straight x over denominator straight x cubed plus straight x squared plus straight x plus 1 end fraction
or             dy over dx space equals space fraction numerator 2 straight x squared plus 1 over denominator straight x squared left parenthesis straight x plus 1 right parenthesis space plus space 1 left parenthesis straight x plus 1 right parenthesis end fraction
or              dy over dx space equals space fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fraction
Separting the variables, we get,
                dy space equals fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fractiondx
Integrating,  integral dy space equals space integral fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fraction dx                    ...(1)
Put    fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fraction space equals fraction numerator straight A over denominator straight x plus 1 end fraction plus fraction numerator Bx plus straight C over denominator straight x squared plus 1 end fraction                     ...(2)
Mutiplying both sides by  open parentheses straight x plus 1 close parentheses space open parentheses straight x squared plus 1 close parentheses comma space space we space get
                         2 straight x squared plus straight x space identical to space straight A space left parenthesis straight x squared plus 1 right parenthesis space plus space left parenthesis Bx plus straight C right parenthesis thin space left parenthesis straight x plus 1 right parenthesis                
or                      2 straight x squared plus straight x space identical to space straight A left parenthesis straight x squared plus 1 right parenthesis space plus Bx left parenthesis straight x plus 1 right parenthesis space plus space straight C left parenthesis straight x plus 1 right parenthesis           ...(3)
Put x + 1 = 0  or  x = -1 in (3)
  therefore               2 - 1 = A(1+1)+0+0   rightwards double arrow space space space space space straight I space equals space 2 straight A space space rightwards double arrow space space straight A space equals space 1 half
            straight x squared right parenthesis space 2 space equals space straight A plus straight B space space space rightwards double arrow space 2 space equals space 1 half plus straight B space space space rightwards double arrow space space space straight B space equals space 3 over 2
          straight x right parenthesis space space 1 space equals space straight B plus straight C space space space space rightwards double arrow space space space space 1 space equals space 3 over 2 plus straight C space space space space rightwards double arrow space space space straight C space equals space minus 1 half
therefore space space space from space left parenthesis 2 right parenthesis comma space space fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fraction space equals space fraction numerator begin display style 1 half end style over denominator straight x plus 1 end fraction plus fraction numerator begin display style 3 over 2 end style straight x minus begin display style 1 half end style over denominator straight x squared plus 1 end fraction
or       fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fraction space identical to space fraction numerator 1 over denominator 2 left parenthesis straight x plus 1 right parenthesis end fraction plus 3 over 2 open parentheses fraction numerator straight x over denominator straight x squared plus 1 end fraction close parentheses space minus space 1 half open parentheses fraction numerator 1 over denominator straight x squared plus 1 end fraction close parentheses
therefore space space space from space left parenthesis 1 right parenthesis comma space space space integral dy space equals space 1 half integral fraction numerator 1 over denominator straight x plus 1 end fraction dx plus 3 over 2 integral fraction numerator straight x over denominator straight x squared plus 1 end fraction dx minus 1 half integral fraction numerator 1 over denominator straight x squared plus 1 end fraction dx
therefore space space space space integral 1 space dy space equals space 1 half integral fraction numerator 1 over denominator straight x plus 1 end fraction dx plus 3 over 4 integral fraction numerator 2 straight x over denominator straight x squared plus 1 end fraction dx minus 1 half integral fraction numerator 1 over denominator straight x squared plus 1 end fraction dx
therefore space space space space space straight y space equals space 1 half log space open vertical bar straight x plus 1 close vertical bar plus 3 over 4 log space left parenthesis straight x squared plus 1 right parenthesis space minus 1 half tan to the power of negative 1 end exponent straight x plus straight c space space space space space space space space space space... left parenthesis 4 right parenthesis
Now   y = 1  when x = 0
therefore space space space space straight I space equals space 1 half log space left parenthesis 1 right parenthesis space plus space 3 over 4 log space 1 space minus space 1 half tan to the power of negative 1 end exponent 0 plus straight c
therefore space space space space straight I space equals space 1 half left parenthesis 0 right parenthesis plus 3 over 4 left parenthesis 0 right parenthesis minus 1 half left parenthesis 0 right parenthesis space plus space straight c space space space space space rightwards double arrow space space space straight c space equals space 1
therefore space space space space from space left parenthesis 4 right parenthesis comma space space straight y space equals space 1 half log space open vertical bar straight x plus 1 close vertical bar plus 3 over 4 log space left parenthesis straight x squared plus 1 right parenthesis minus 1 half tan to the power of negative 1 end exponent straight x plus 1





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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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 Multiple Choice QuestionsShort Answer Type

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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 Multiple Choice QuestionsLong Answer Type

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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