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Vector Algebra

Question
CBSEENMA12033047
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Solve 3ex tan y dx + (1 + ex) sec2 y dy = 0. 

Solution
The given differential equation is
                   3ex tan y dx + (1 + ex) sec2 y dy = 0
     or       left parenthesis 1 plus straight e to the power of straight x right parenthesis space sec squared straight y space dy space equals space minus space 3 space straight e to the power of straight x space tan space straight y space dx
or            fraction numerator sec squared straight y over denominator tan space straight y end fraction dy space equals space minus 3 fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of straight x end fraction dx
Integrating, 
                            integral fraction numerator sec squared straight y over denominator tan space straight y end fraction dy space equals space minus 3 space integral fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of straight x end fraction dx
therefore space space space space log space open vertical bar tan space straight y close vertical bar space equals space minus 3 space log space left parenthesis 1 plus straight e to the power of straight x right parenthesis space plus space straight c comma which is required solution..

Some More Questions From Vector Algebra Chapter

Determine order and degree (if defined) of differential equations:
y' + y = ex

Determine order and degree (if defined) of differential equations:
y'' + (y')2 + 2 y = 0

Determine order and degree (if defined) of differential equations:
y″ + 2y′ + sin y = 0

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