Question

Solve the following differential equation:
$open parentheses straight x squared straight y plus straight x squared close parentheses space dx space plus space left parenthesis straight x space straight y squared minus straight y squared right parenthesis space dy space equals space 0$

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Solution

The given differential equation is

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

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

Separating the variables, we get,

fraction numerator straight y squared over denominator straight y plus 1 end fraction dy space equals space minus fraction numerator straight x squared over denominator straight x minus 1 end fraction dxaa

Integrating,

integral fraction numerator straight y squared over denominator straight y plus 1 end fraction dy space equals space minus integral fraction numerator straight x squared over denominator straight x minus 1 end fraction dx

therefore space space space space integral open parentheses straight y minus 1 plus fraction numerator 1 over denominator straight y plus 1 end fraction close parentheses space dy space equals space minus integral open parentheses straight x plus 1 plus fraction numerator 1 over denominator straight x minus 1 end fraction close parentheses dx therefore space space space straight y squared over 2 minus straight y plus log space open vertical bar straight y plus 1 close vertical bar space equals space open square brackets straight x squared over 2 plus straight x plus log space open vertical bar straight x minus 1 close vertical bar close square brackets space plus space straight c

which is required solution.

For Daily Free Study Material Join wiredfaculty