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Vector Algebra

Question
CBSEENMA12033048
Wired Faculty App

 Solve 4 ex tan y dx + 3(1 + ex ) sec2 y dy = 0. 

Solution
The given differential equation is 4 ex tan y dx + 3(1 + ex ) sec2 y dy = 0.
therefore space space 3 space left parenthesis 1 plus straight e to the power of straight x right parenthesis space sec squared straight y space dy space equals negative space 4 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 4 over 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
therefore space space space space integral fraction numerator sec squared straight y over denominator tan space straight y end fraction dy space equals space minus 4 over 3 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 log space open vertical bar tan space straight y close vertical bar space equals negative 4 over 3 space log space left parenthesis 1 plus straight e to the power of straight x right parenthesis space plus space log space straight c therefore space log space open vertical bar tan space straight y close vertical bar space equals space log space left parenthesis 1 plus straight e to the power of straight x right parenthesis to the power of fraction numerator negative 4 over denominator 3 end fraction end exponent space plus space log space straight c. space space space space space therefore space space space log space open vertical bar tan space straight y close vertical bar space equals space log space fraction numerator straight c over denominator left parenthesis 1 plus straight e to the power of straight x right parenthesis to the power of begin display style 4 over 3 end style end exponent end fraction therefore space space space space space open vertical bar tan space straight y close vertical bar space equals fraction numerator straight c over denominator left parenthesis 1 plus straight e to the power of straight x right parenthesis to the power of begin display style 4 over 3 end style end exponent end fraction space is space the space required space solution. space
             

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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