Find a one-parameter family of solutions of the following differential equation, indicating carefully the interval in which the solutions are valid:
$dy plus left parenthesis straight x plus 1 right parenthesis space left parenthesis straight y plus 1 right parenthesis space dx space equals space 0$

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Solution

The given differential equation is

dy plus left parenthesis straight x plus 1 right parenthesis space left parenthesis straight y plus 1 right parenthesis space dx space equals space 0 space space space space space space space space space or space space space space dy space equals space minus left parenthesis straight x plus 1 right parenthesis space left parenthesis straight y plus 1 right parenthesis space dx

therefore space space space space space space fraction numerator 1 over denominator straight y plus 1 end fraction dy space equals space minus left parenthesis straight x plus 1 right parenthesis space dx

Integrating,

integral fraction numerator 1 over denominator straight y plus 1 end fraction dy space equals space integral left parenthesis straight x plus 1 right parenthesis space dx

therefore space space space log space open vertical bar straight y plus 1 close vertical bar space equals space minus open parentheses straight x squared over 2 plus straight x close parentheses space plus space straight c comma space space space straight x space element of space straight R space is space the space solution. space

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