Question:
Solution of \(\dfrac{dy}{dx} – y = 1\), y(0) = 1 is given by
\(\dfrac{dy}{dx} – y = 1\), y(0) = 1 का हल क्या है?
Options:
- \(y = 2e^x – 1\)
- \(xy = -e^x\)
- \(xy = -e^{-x}\)
- xy = −1
- \(y = 2e^x – 1\)
- \(xy = -e^x\)
- \(xy = -e^{-x}\)
- xy = −1
Correct Answer: a)
Explanation:
\(\dfrac{dy}{dx} – y = 1 \implies \text{IF} = e^{-x}\).
\(y e^{-x} = \int e^{-x}\,dx = -e^{-x} + C \implies y = Ce^x – 1\).
Using \(y(0) = 1\): \(1 = C – 1 \implies C = 2\).
Solution is \(y = 2e^x – 1\).
\(y e^{-x} = -e^{-x} + C \implies y = Ce^x – 1\)।
\(y(0) = 1 \implies C = 2\)।
अतः हल \(y = 2e^x – 1\) है।