Question

Solve the differential equation:
$left parenthesis straight y plus straight x right parenthesis space dy over dx space equals space straight y minus straight x.$

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Solution

The given differential equation is

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

therefore space space space space space space space dy over dx space equals space fraction numerator straight y minus straight x over denominator straight y plus straight x end fraction

...(1)
Put y = v x so that

dy over dx space equals space straight v plus straight x dv over dx

therefore

from (1),

straight v plus straight x dv over dx space equals space fraction numerator vx minus straight x over denominator vx plus straight x end fraction

therefore space space space space straight v plus straight x dv over dx space equals space fraction numerator straight v minus 1 over denominator straight v plus 1 end fraction space space space space space space space space space space space space space or space space space space space space space straight x dv over dx equals fraction numerator straight v minus 1 over denominator straight v plus 1 end fraction minus straight v

straight x dv over dx space equals fraction numerator straight v minus 1 minus straight v squared minus straight v over denominator straight v plus 1 end fraction space space space space or space space space space straight x dv over dx space equals space minus fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction

Separating the variables, we get,

fraction numerator 1 plus straight v over denominator 1 plus straight v squared end fraction dv space equals space minus 1 over straight x dx

Integrating,

integral fraction numerator 1 plus straight v over denominator 1 plus straight v squared end fraction dv equals space minus integral 1 over straight x dx

therefore space space space space integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv plus 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space minus integral 1 over straight x dx

therefore space space space space tan to the power of negative 1 end exponent straight v plus 1 half log space left parenthesis 1 plus straight v squared right parenthesis space equals space minus log space open vertical bar straight x close vertical bar space plus space straight c subscript 1 therefore space space space space space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus 1 half log open parentheses 1 plus straight y squared over straight x squared close parentheses space equals space minus log space open vertical bar straight x close vertical bar plus straight c subscript 1 therefore space space space space space 2 space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus log space open parentheses fraction numerator straight x squared plus straight y squared over denominator straight x squared end fraction close parentheses space equals space minus log open vertical bar straight x close vertical bar squared plus 2 straight c subscript 1 therefore space space space space space 2 space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus log space open parentheses fraction numerator straight x squared plus straight y squared over denominator straight x squared end fraction close parentheses space equals space minus log space straight x squared plus straight c therefore space space space 2 space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus log space open parentheses fraction numerator straight x squared plus straight y squared over denominator straight x squared end fraction close parentheses plus log space straight x squared space equals space straight c therefore space space space 2 space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus log space open parentheses fraction numerator straight x squared plus straight y squared over denominator straight x squared end fraction cross times space straight x squared close parentheses space equals space straight c therefore space space space 2 space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus log space open parentheses straight x squared plus straight y squared close parentheses space equals space straight c

which is required solution.

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Differential Equations

Solve the differential equation:
$left parenthesis straight y plus straight x right parenthesis space dy over dx space equals space straight y minus straight x.$

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