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Find the Solution of the differential equation \(\frac{{dx}}{{dy}} = \frac{{1 + {x^2}}}{y}\) with condition that y = 1, x = 0.1. \(y = {e^{ - {{\tan }^{ - 1}}x}}\)2. \(y = - {e^{ - {{\tan }^{ - 1}}x}}\)3. \(y = {e^{{{\tan }^{ - 1}}x}}\)4. \(y = - {e^{{{\tan }^{ - 1}}x}}\) |
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Answer» Correct Answer - Option 3 : \(y = {e^{{{\tan }^{ - 1}}x}}\) 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: \(\frac{{dx}}{{dy}} = \frac{{1 + {x^2}}}{y}\) \(\frac{{dx}}{{1 + {x^2}}} = \frac{{dy}}{y}\) Integrate both side \(\int \frac{{dx}}{{1 + {x^2}}} = \int \frac{{dy}}{y}\) \({\tan ^{ - 1}}x = \ln y + \ln c\) \({\tan ^{ - 1}}x = \ln yc\) \(yc = {e^{{{\tan }^{ - 1}}x}}\) Put x = 0, y = 1 \(c = {e^{{{\tan }^{ - 1}}0}} = 1\) \(y = {e^{{{\tan }^{ - 1}}x}}\) |
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