If curve represented by differential equation \(x\displaystyle\frac{dy}{dx}+y=e^{x}\) passes through (1, 1), then y(−1) is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{e^{2}-(e+1)}{e}\)
The left side is the derivative of a product, since \(\displaystyle\frac{d}{dx}(xy) = x\frac{dy}{dx} + y\). So the equation is
\( \displaystyle \frac{d}{dx}(xy) = e^{x} \)
Integrating:
\(xy = e^{x} + C\)
The curve passes through (1, 1):
\(1 = e + C\)
\(C = 1 - e\)
So the curve is \(xy = e^{x} + 1 - e\). At x = −1:
\(-y = e^{-1} + 1 - e\)
\( \displaystyle y = e - 1 - \frac{1}{e} \)
\( \displaystyle = \frac{e^{2} - e - 1}{e} = \frac{e^{2}-(e+1)}{e} \)