Question

Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e^−x + 2

Answer 1

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.