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The solution of the DE \(\frac{dy}{dx}\) = ex+y isA. ex + ey = cB. ex - e-y = cC. ex + e-y = cD. None of these |
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Answer» Given, \(\frac{dy}{dx}\) = ex+y \(\frac{dy}{dx}\) = exey e-y dy = ex dx On integrating on both sides, we get - e-y + c1 ex +c2 e-y + ex = c Conclusion: Therefore, e-y + ex = c is the solution of \(\frac{dy}{dx} \) ex+y |
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