Question

Solve:
$open parentheses straight e to the power of straight x plus 1 close parentheses space straight y space dy space plus space left parenthesis straight y plus 1 right parenthesis space straight e to the power of straight x space dx space equals space 0$

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Solution

The given differential equation is

open parentheses straight e to the power of straight x plus 1 close parentheses space straight y space dy space plus space left parenthesis straight y plus 1 right parenthesis space straight e to the power of straight x space dx space equals space 0

open parentheses straight e to the power of straight x plus 1 close parentheses straight y space dy space equals space minus left parenthesis straight y plus 1 right parenthesis space straight e to the power of straight x dx

fraction numerator straight y over denominator straight y plus 1 end fraction dy space equals space minus fraction numerator straight e to the power of straight x over denominator straight e to the power of straight x plus 1 end fraction dx

Integrating,

integral fraction numerator straight y over denominator straight y plus 1 end fraction dy space equals space minus integral fraction numerator straight e to the power of straight x over denominator straight e to the power of straight x plus 1 end fraction dx

therefore space space space space integral open parentheses 1 minus fraction numerator 1 over denominator straight y plus 1 end fraction close parentheses space dy space equals space minus integral fraction numerator straight e to the power of straight x over denominator straight e to the power of straight x plus 1 end fraction dx therefore space space straight y minus log space open vertical bar straight y plus 1 close vertical bar space equals space minus space log space left parenthesis straight e to the power of straight x plus 1 right parenthesis space plus straight c

which is the required solution.

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