Differential Equations

Question

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
$straight y space dx space plus space straight x space open parentheses log space straight y over straight x close parentheses space dy space minus space 2 space straight x space dy space equals space 0$

Solution

The given differential equation is
$straight y space dx plus straight x space log space open parentheses straight y over straight x close parentheses space dy space minus 2 straight x space dy space equals space 0$
$therefore space space straight x space open parentheses 2 minus log space straight y over straight x close parentheses space dy space equals space straight y space dx therefore space space space space space space space space space space space space space space space space space space space space space dy over dx equals fraction numerator begin display style straight y over straight x end style over denominator 2 minus log space begin display style straight y over straight x end style end fraction$
Put y = v x so that $dy over dx equals straight v plus dv over dx$
$therefore space space space space space space space space space straight v plus straight x dv over dx equals fraction numerator straight v over denominator 2 minus log space straight v end fraction therefore space space space space space space space space straight x space dv over dx space equals space fraction numerator straight v over denominator 2 minus log space straight v end fraction minus straight v therefore space space space space straight x dv over dx equals fraction numerator straight v minus 2 straight v plus straight v space log space straight v over denominator 2 minus log space straight v end fraction therefore space space space space space straight x dv over dx space equals space fraction numerator negative straight v plus straight v space log space straight v over denominator 2 minus space log space straight v end fraction$
$therefore space space space space fraction numerator 2 minus space log space straight v over denominator negative straight v plus straight v space log space straight v end fraction dv space equals space 1 over straight x dx therefore space space space space integral fraction numerator 1 minus log space left parenthesis straight v minus 1 right parenthesis over denominator straight v left parenthesis log space straight v minus 1 right parenthesis space end fraction dv space equals space integral 1 over straight x dx therefore space space space space integral fraction numerator 1 over denominator straight v left parenthesis log space straight v minus 1 right parenthesis end fraction dv space minus space integral 1 over straight v dv space equals space integral 1 over straight x dx therefore space space space integral fraction numerator begin display style 1 over straight v end style over denominator log space straight v minus 1 end fraction dv space minus space integral 1 over straight v dv space equals space integral 1 over straight x dx therefore space space space log space open vertical bar log space space straight v minus 1 close vertical bar space minus space log space open vertical bar straight v close vertical bar space equals space log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight c close vertical bar therefore space space space log space open vertical bar fraction numerator log space straight v minus 1 over denominator straight v end fraction close vertical bar space equals space log space left parenthesis open vertical bar cx close vertical bar right parenthesis therefore space space space space log space open vertical bar fraction numerator log space begin display style straight y over straight x end style minus 1 over denominator begin display style straight y over straight x end style end fraction close vertical bar space equals space log space left parenthesis open vertical bar straight c space straight x close vertical bar right parenthesis therefore space space space space space space fraction numerator log space begin display style straight y over straight x end style minus 1 over denominator begin display style straight y over straight x end style end fraction space equals straight c space straight x therefore space space space log space straight y over straight x minus 1 space equals space straight c space straight y space is space the space required space solution.$

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