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Find the solution of the differential equation \(\frac{{di}}{{dt}} = 1 + \frac{i}{t} + {\left( {\frac{i}{t}} \right)^2}\)1. \({\tan ^{ - 1}}i = \ln ct\)2. \({\tan ^{ - 1}}\frac{i}{t} = \ln ct\)3. \({\cot ^{ - 1}}i = \ln ct\)4. \({\cot ^{ - 1}}\frac{i}{t} = \ln ct\) |
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Answer» Correct Answer - Option 2 : \({\tan ^{ - 1}}\frac{i}{t} = \ln ct\) Concept: Homogenous equations are of the form \(\frac{{di}}{{dt}} = \frac{{f\left( {x,y} \right)}}{{\emptyset \left( {x,y} \right)}}\) Where \(f\left( {x,y} \right)and\;\emptyset \left( {x,y} \right)\) Homogenous functions of the same degree in x and y. To solve a homogenous equation
Calculation: \(\frac{{di}}{{dt}} = 1 + \frac{i}{t} + {\left( {\frac{i}{t}} \right)^2}\) Put i = vt, then \(\frac{{di}}{{dt}} = v + t\frac{{di}}{{dt}}\) \(v + t\frac{{dv}}{{dt}} = 1 + v + {v^2}\) \(\frac{{dv}}{{1 + {v^2}}} = \frac{{dt}}{t}\) By integrating both sides we get \(\smallint \frac{{dv}}{{1 + {v^2}}} = \smallint \frac{{dt}}{t}\) \({\tan ^{ - 1}}v = \ln ct\) Put v = 1/t in the above equation we get, \({\tan ^{ - 1}}\frac{i}{t} = \ln ct\) |
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