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Find the solution of different equation \(\left( {y + x\frac{{dy}}{{dx}}} \right) = a\left( {{y^2} - 2\frac{{dy}}{{dx}}} \right)\)1. \(\frac{{ay + 1}}{y} = c\left[ {x + 2a} \right]\)2. \(\frac{y}{{ay + 1}} = c\left[ {x + 2a} \right]\)3. \(\frac{{ay - 1}}{y} = c\left[ {x + 2a} \right]\)4. \(\frac{y}{{ay - 1}} = c\left[ {x + 2a} \right]\) |
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Answer» Correct Answer - Option 3 : \(\frac{{ay - 1}}{y} = c\left[ {x + 2a} \right]\) Concept: If in an equation it is possible to collect all function of x and dx on one side and all function of y and dy on the other side, then the variables are said to be separable. Thus the general form of such an equation is \(f\left( y \right)dy = \emptyset \left( x \right)dx\) Integrating both sides, we get \(\smallint f\left( y \right)dy = \smallint \emptyset \left( x \right)dx + c\) as its solution. Calculation: \(\left( {y + x\frac{{dy}}{{dx}}} \right) = a\left( {{y^2} - 2\frac{{dy}}{{dx}}} \right)\) \(ydx + xdy = a{y^2}dx - 2ady\) \(\left( {x + 2a} \right)dy = \left( {a{y^2} - y} \right)dx\) \(\frac{{dy}}{{a{y^2} - y}} = \frac{{dx}}{{x + 2a}}\) \(\left( {\frac{a}{{y - 1}} - \frac{1}{y}} \right)dy = \frac{{dx}}{{x + 2a}}\) Integrate both side \(\smallint \left( {\frac{a}{{y - 1}} - \frac{1}{y}} \right)dy = \smallint \frac{{dx}}{{x + 2a}}\) \(\ln \left( {ay - 1} \right) - \ln y = \ln \left( {x + 2a} \right) + \ln c\) \(\ln \frac{{ay - 1}}{y} = \ln c\left[ {x + 2a} \right]\) \(\frac{{ay - 1}}{y} = \;c\left[ {x + 2a} \right]\) |
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