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Differential Equations

Question
CBSEENMA12033081

Find the equation of the curve passing through the point open parentheses 0 comma space straight pi over 4 close parentheses whose  differential equation is sin x cos y dx + cos x sin y dy = 0.

Solution

The given differential equation is
                        sin x cos y dx + cos x sin y dy = 0
  or          sin x cos y dx =  - cos x sin y dy
therefore space space space space space space space space space fraction numerator negative sin space straight y over denominator cos space straight y end fraction dy space equals space fraction numerator sin space straight x over denominator cos space straight x end fraction dx
Integrating integral fraction numerator negative sin space straight y over denominator cos space straight y end fraction dy space equals space fraction numerator sin space straight x over denominator cos space straight x end fraction dx
therefore space space log space open vertical bar cos space straight y close vertical bar space plus space log space open vertical bar cos space straight x close vertical bar space equals space log space straight A space space space space space space space rightwards double arrow space space space space space log space open vertical bar cosx space cosy close vertical bar space equals space log space straight A
therefore space space space open vertical bar cosx space cosy close vertical bar space equals space straight A space space space space space space space space space space space space space space space rightwards double arrow space space space space cos space straight x space cos space straight y space equals space 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 space space... left parenthesis 1 right parenthesis
Since the curve passes through open parentheses 0 comma space straight pi over 4 close parentheses
therefore space space space cos space 0 space cos space straight pi over 4 space equals straight c space space space space space space rightwards double arrow space space space 1 space cross times space fraction numerator 1 over denominator square root of 2 end fraction space equals space straight c space space space space space rightwards double arrow space space space straight c space equals space fraction numerator 1 over denominator square root of 2 end fraction
therefore space space from space left parenthesis 1 right parenthesis comma space cosx space cos space straight y space space equals space fraction numerator 1 over denominator square root of 2 end fraction comma space space which space is space required space equation.

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