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Question
Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−x + 2
Solution
We have,
\[y = e^{- x} + 2..............(1)\]
Differentiating both sides of (1) with respect to x, we get
\[\frac{dy}{dx} = - e^{- x} \]
\[ \Rightarrow \frac{dy}{dx} = - \left( y - 2 \right) ..............\left[\text{Using (1)}\right]\]
\[ \Rightarrow \frac{dy}{dx} + y = 2 \]
It is the given differential equation.
\[y = e^{- x} + 2\] satisfies the given differential equation; hence, it is a solution.
Also, when \[x = 0, y = e^0 + 2 = 1 + 2 = 3,\text{ i.e. }y(0) = 3\]
Hence, \[y = e^{- x} + 2\] is the solution to the given initial value problem.
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