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Differential Equations

Question
CBSEENMA12033119
Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
2 space straight y space straight e to the power of straight x over 4 end exponent space dx plus open parentheses straight y minus 2 space straight x space straight e to the power of straight x over straight y end exponent close parentheses space dy space equals space 0

Solution
The given differential equation is
                          2 ye to the power of straight x over straight y end exponent dx plus open parentheses straight y minus 2 xe to the power of straight x over straight y end exponent close parentheses dy space equals space 0
therefore space space space space space space space space open parentheses straight y minus 2 xe to the power of straight x over straight y end exponent close parentheses space dy space equals space minus 2 space straight y space straight e to the power of straight x over straight y end exponent dx therefore space space space space space space dx over dy space equals negative fraction numerator straight y minus 2 xe to the power of begin display style straight x over straight y end style end exponent over denominator 2 ye to the power of begin display style straight x over straight y end style end exponent end fraction therefore space space space space space space space dx over dy space equals space fraction numerator straight y open parentheses 2 begin display style straight x over straight y end style straight e to the power of begin display style straight x over straight y end style end exponent minus 1 close parentheses over denominator 2 straight y space. straight e to the power of begin display style straight x over straight y end style end exponent end fraction space space space space space space space or space space space space space space space space dx over dy space equals space fraction numerator 2 begin display style straight x over straight y end style straight e to the power of begin display style straight x over straight y end style end exponent minus 1 over denominator 2 straight e to the power of begin display style straight x over straight y end style end exponent end fraction
Put x = v y so that  dx over dy space equals space straight v plus straight y dv over dy
therefore space space space straight v plus straight y dv over dy space equals fraction numerator 2 ve to the power of straight v minus 1 over denominator 2 straight e to the power of straight v end fraction. space space space space space space therefore space space straight y space dv over dy space equals space fraction numerator 2 straight v space straight e to the power of straight v minus 1 over denominator 2 straight e to the power of straight v end fraction minus straight v therefore space space space space space straight y dv over dy space equals space fraction numerator 2 ve to the power of straight v minus 1 minus 2 ve to the power of straight v over denominator 2 straight e to the power of straight v end fraction therefore space space space straight y dv over dy space equals space minus fraction numerator 1 over denominator 2 straight e to the power of straight v end fraction therefore space space space space space straight e to the power of straight v space dv space equals space minus fraction numerator 1 over denominator 2 space straight y end fraction dy therefore space space space 2 space integral straight e to the power of straight v space dv space equals space minus integral 1 over straight y dy therefore space space space space space space space space 2 space straight e to the power of straight v space equals space minus log space open vertical bar straight y close vertical bar space plus space straight c therefore space space space space space space 2 space straight e to the power of straight x over straight y end exponent space equals space minus log space open vertical bar straight y close vertical bar plus straight c space is space the space required space solution.

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