Question

The solution $ \frac{dy}{dx}+y={{e}^{x}} $ is

• A. $ 2y={{e}^{2x}}+c $
• B. s $ 2y{{e}^{x}}={{e}^{2}}+c $
• C. $ 2y{{e}^{x}}={{e}^{2x}}+c $
• D. $ 2y{{e}^{2x}}=2{{e}^{x}}+c $

Answer 1

Answer: (C)

Given, differential equation is
$ \frac{dy}{x}+y={{e}^{x}}, $ which is of the form
$ \frac{dy}{dx}+Py=Q $
Here, $ P=1,\,\,Q={{e}^{x}} $
$ \therefore $ $ IF={{e}^{\int{Pdx}}}={{e}^{\int{1dx}}}={{e}^{x}} $
Now, solution is $ y.IF\,=\,\int{Q.\,IF\,\,dx} $
$ \Rightarrow $ $ y{{e}^{x}}=\int{{{e}^{2x}}\,\,dx} $
$ \Rightarrow $ $ y{{e}^{x}}=\frac{{{e}^{2x}}}{2}+c $
$ \Rightarrow $ $ 2y{{e}^{x}}={{e}^{2x}}+c $