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Solution of the differential equation $$ydx + \left( {x + {x^2}y} \right)dy = 0$$ is

JEE · Math · previous-year question

  1. A.$$log$$ $$y=Cx$$
  2. B.$$ - {1 \over {xy}} + \log y = C$$correct
  3. C.$${1 \over {xy}} + \log y = C$$
  4. D.$$ - {1 \over {xy}} = C$$

Answer

B. $$ - {1 \over {xy}} + \log y = C$$

Explanation

$$ydx + \left( {x + {x^2}y} \right)dy = 0$$ $$ \Rightarrow {{dx} \over {dy}} = - {x \over y} - {x^2}$$ $$ \Rightarrow {{dx} \over {dy}} + {x \over y} = - {x^2},$$ It is Bernoullis form. Divide by $${x^2}$$ $${x^{ - 2}}{{dx} \over {dy}} + {x^{ - 1}}\left( {{1 \over y}} \right) = - 1.$$ put $${x^{ - 1}} = t,\,\, - {x^{ - 2}}{{dx} \over {dy}} = {{dt} \over {dy}}$$ We get, $$ - {{dt} \over {dy}} + t\left( {{1 \over y}} \right) = - 1$$ $$ \Rightarrow {{dt} \over {dy}} - \left( {{1 \over y}} \right)t = 1$$ It is linear in $$t.$$ Integrating factor $$ = {e^{\int { - {1 \over y}dy} }} = {e^{ - \log y}} = {y^{ - 1}}$$ $$\therefore$$ Solution is $$t\left( {{y^{ - 1}}} \right) = \int {\left( {{y^{ - 1}}} \right)} dy + c$$ $$ \Rightarrow {1 \over x}.{1 \over y} = \log y + c$$ $$ \Rightarrow \log y - {1 \over {xy}} = c$$

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