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Differential Equations

Long Answer Type

201. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

1 over straight x cos straight y over straight x dx minus open parentheses straight x over straight y sin straight y over straight x plus cos straight y over straight x close parentheses dy space equals 0

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202. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

2 space straight y space straight e to the power of straight x over 4 end exponent space dx plus open parentheses straight y minus 2 space straight x space straight e to the power of straight x over straight y end exponent close parentheses space dy space equals space 0

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203. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

straight y space dx space plus space straight x space open parentheses log space straight y over straight x close parentheses space dy space minus space 2 space straight x space dy space equals space 0

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204. Show that the given differential equation is homogeneous and solve it:

left parenthesis straight x squared plus xy right parenthesis space dy space equals space left parenthesis straight x squared plus straight y squared right parenthesis space dx

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Short Answer Type

205. Show that the given differential equation is homogeneous and solve it:

straight y apostrophe space equals space fraction numerator straight x plus straight y over denominator straight x end fraction

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206. Show that the given differential equation is homogeneous and solve it:
(x-y) dy - (x+y) dx = 0

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207. Show that the given differential equation is homogeneous and solve it:

open parentheses straight x squared minus straight y squared close parentheses space dx space plus space 2 xy space dy space equals space 0

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Long Answer Type

208. Show that the given differential equation is homogeneous and solve it:

straight x squared dy over dx space equals space straight x squared minus 2 straight y squared plus straight x space straight y

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209. Show that the given differential equation is homogeneous and solve it:
$open curly brackets straight x space cos space open parentheses straight y over straight x close parentheses plus straight y space sin space open parentheses straight y over straight x close parentheses close curly brackets straight y space dx space equals space open curly brackets straight y space sin space open parentheses straight y over straight x close parentheses minus straight x space cos space open parentheses straight y over straight x close parentheses close curly brackets space straight x space dy$

The given differential equation is
$open curly brackets straight x space cos space open parentheses straight y over straight x close parentheses space plus space straight y space sin space open parentheses straight y over straight x close parentheses close curly brackets straight y space dy space equals space open curly brackets straight y space sin space open parentheses straight y over straight x close parentheses minus straight x space cos space open parentheses straight y over straight x close parentheses close curly brackets straight x space dy$
or $dy over dx space equals space fraction numerator straight x space straight y space open curly brackets cos space begin display style straight y over straight x end style plus begin display style straight y over straight x end style sin space begin display style straight y over straight x end style close curly brackets over denominator straight x squared open curly brackets begin display style straight y over straight x end style sin begin display style straight y over straight x end style minus cos space begin display style straight y over straight x end style close curly brackets end fraction$
or    $dy over dx space equals space fraction numerator begin display style straight y over straight x end style open curly brackets cos space begin display style straight y over straight x end style plus begin display style straight y over straight x end style sin begin display style straight y over straight x end style close curly brackets over denominator begin display style straight y over straight x end style sin space begin display style straight y over straight x end style minus cos begin display style straight y over straight x end style end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$
Since R.H.S. of (1) is of the form $straight F open parentheses straight y over straight x close parentheses comma$ therefore, it is a homogeneous differential equation.
Put y= v x so that $dy over dx space equals space straight v plus straight x space dv over dx$
$therefore space space space space left parenthesis 1 right parenthesis space space becomes comma space space space straight v plus straight x dv over dx space equals space fraction numerator straight v left parenthesis cos space straight v space plus space straight v space sin space straight v right parenthesis space over denominator straight v space sin space straight v space minus space cos space straight v end fraction$

or    $straight x dv over dx space equals space fraction numerator straight v space cosv plus straight v squared space sin space straight v over denominator straight v space sin space straight v space minus space cos space straight v end fraction minus straight v$

or $straight x dv over dx space equals space fraction numerator 2 space straight v space cos space straight v over denominator straight v space sin space straight v space minus space cos space straight v end fraction$

Separating the variables,
   $open parentheses fraction numerator straight v space sin space straight v space minus space cos space straight v over denominator straight v space cos space straight v end fraction close parentheses space dv space equals space 2 over straight x dx space space or space space space open parentheses tan space straight v minus space 1 over straight v close parentheses dv space equals space 2 over straight x dx$
Integrating, $integral open parentheses tan space straight v minus 1 over straight v close parentheses space dv space equals space 2 space integral 1 over straight x dx$
$therefore space space space minus log space open vertical bar cos space straight v close vertical bar minus log space open vertical bar straight v close vertical bar space equals space 2 space log space open vertical bar straight x close vertical bar plus straight c rightwards double arrow space space space space log space open vertical bar straight v space cos space straight v close vertical bar space plus space 2 space log open vertical bar straight x close vertical bar space equals space minus straight c rightwards double arrow space space space log space left parenthesis open vertical bar straight v space cos space straight v close vertical bar space open vertical bar straight x squared close vertical bar right parenthesis space equals space minus straight c rightwards double arrow space space space space space open vertical bar straight v space cos space straight v close vertical bar space open vertical bar straight x close vertical bar squared space equals space straight e to the power of negative straight c end exponent rightwards double arrow space space straight x squared straight v space cos space straight v space equals space plus-or-minus straight e to the power of negative straight c end exponent rightwards double arrow space space space space straight x squared straight y over straight x cos space open parentheses straight y over straight x close parentheses space equals straight A comma space space where space straight A space space equals space plus-or-minus space straight e to the power of negative straight c end exponent rightwards double arrow space space space straight x space straight y space cos space straight y over straight x space equals straight A comma space which space is space the space required space general space solution. space$

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210. Solve

straight x space dy space minus space straight y space dx space equals space square root of straight x squared plus straight y squared end root space dx.

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