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Solution of given differential equation \(dy = \left( {{e^y} + {x^2}{e^y}} \right)dx\) is1. \(y = x + {\frac{x}{3}^3} + c\)2. \({e^{ - y}} = - x - {\frac{x}{3}^3} + c\)3. \({e^{ - y}} = x + {\frac{x}{3}^3} + c\)4. \(y = - x - {\frac{x}{3}^3} + c\) |
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Answer» Correct Answer - Option 2 : \({e^{ - y}} = - x - {\frac{x}{3}^3} + c\) 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: \(dy = \left( {{e^y} + {x^2}{e^y}} \right)dx\) \(dy = {e^y}\left( {1 + {x^2}} \right)dx\) \({e^{ - y}}dy = \left( {1 + {x^2}} \right)dx\) Integrate both side \(\smallint {e^{ - y}}dy = \smallint \left( {1 + {x^2}} \right)dx\) \( - {e^{ - y}} = x + {\frac{x}{3}^3} + c_1\) \({e^{ - y}} = - x - {\frac{x}{3}^3} + c\) where c = - c1 |
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