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Differential Equations

Short 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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Long 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

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186. For the following differential equation, given below, find particular solution satisfying the given condition:
$straight x open parentheses straight x squared minus 1 close parentheses space dy over dx space equals space 1 space semicolon space space straight y space equals space 0 space space when space straight x space equals space 2$

The given differential equation is

straight x open parentheses straight x squared minus 1 close parentheses dy over dx space equals space 1 space space space or space space dy over dx space equals space fraction numerator 1 over denominator straight x left parenthesis straight x squared minus 1 right parenthesis end fraction

Separating the variables, we get.

dy space equals space fraction numerator 1 over denominator straight x left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x plus 1 right parenthesis end fraction dx

Integrating,

integral space dy space equals space integral fraction numerator 1 over denominator straight x space left parenthesis straight x minus 1 right parenthesis space left parenthesis straight x plus 1 right parenthesis end fraction space dx

integral dy space equals space integral open square brackets fraction numerator 1 over denominator straight x left parenthesis 0 minus 1 right parenthesis thin space left parenthesis 0 plus 1 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis 1 right parenthesis space left parenthesis straight x minus 1 right parenthesis thin space left parenthesis 1 plus 1 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis negative 1 right parenthesis thin space left parenthesis negative 1 minus 1 right parenthesis space left parenthesis straight x plus 1 right parenthesis end fraction close square brackets dx therefore space space space space integral space 1 space dy space equals space integral open square brackets negative 1 over straight x plus fraction numerator 1 over denominator 2 left parenthesis straight x minus 1 right parenthesis end fraction plus fraction numerator 1 over denominator 2 left parenthesis straight x plus 1 right parenthesis end fraction close square brackets dx therefore space space space straight y space equals space minus log space open vertical bar straight x close vertical bar plus 1 half space log space open vertical bar straight x minus 1 close vertical bar space plus space 1 half log space open vertical bar straight x plus 1 close vertical bar plus straight c space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Now y = 0 when x = 2

therefore space space space space 0 space equals space minus log space 2 space plus 1 half log space 1 space plus space 1 half log space 3 space plus straight c rightwards double arrow space space space space 0 space equals space minus space log space 2 plus 1 half cross times 0 plus 1 half log space 3 space plus straight c rightwards double arrow space 0 space equals space minus 1 half log space 4 plus 1 half space log 3 space plus straight c rightwards double arrow space space space straight c space equals space minus 1 half left parenthesis log space 3 space minus space log space 4 right parenthesis space space space space rightwards double arrow space space space straight c space equals space minus 1 half log open parentheses 3 over 4 close parentheses therefore space space from space left parenthesis 1 right parenthesis comma space space space space space space space straight y space equals space minus log space open vertical bar straight x close vertical bar plus 1 half log space open vertical bar straight x minus 1 close vertical bar plus 1 half log space open vertical bar straight x plus 1 close vertical bar minus 1 half log space open parentheses 3 over 4 close parentheses or space space space space straight y space equals space 1 half open square brackets log space open vertical bar straight x minus 1 close vertical bar plus log space open vertical bar straight x plus 1 close vertical bar minus 2 space log open vertical bar straight x close vertical bar space close square brackets minus 1 half log space open parentheses 3 over 4 close parentheses or space space space space straight y space equals space 1 half open square brackets log space open vertical bar space left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x plus 1 right parenthesis close vertical bar minus log space open vertical bar straight x close vertical bar squared close square brackets space minus space 1 half log space 3 over 4 or space space space straight y space equals space 1 half log space open vertical bar fraction numerator straight x squared minus 1 over denominator straight x squared end fraction close vertical bar space minus space 1 half log space 3 over 4 space is space the space required space solution. space

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Short 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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Long 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:
$straight x squared dy over dx space equals space 2 xy plus straight y squared$

74 Views