Question

Differential equation \[\frac{d^2 y}{d x^2} - y = 0, y \left( 0 \right) = 2, y' \left( 0 \right) = 0\] Function y = e^x + e^−x

Answer 1

We have,

\[y = e^x + e^{−x}............(1)\]

Differentiating both sides of (1) with respect to x, we get

\[\frac{dy}{dx} = e^x - e^{- x}..........(2)\]

Differentiating both sides of (2) with respect to x, we get

\[\frac{d^2 y}{d x^2} = e^x + e^{- x} \]

\[ \Rightarrow \frac{d^2 y}{d x^2} = y ..........\left[\text{Using (1)}\right]\]

\[\frac{d^2 y}{d x^2} - y = 0\]

It is the given differential equation.

Therefore, y = ex + e−x satisfies the given differential equation.

Also, when \[x = 0; y = e^0 + e^0 = 1 + 1,\text{ i.e. }y(0) = 2\]
And, when \[x = 0; y_1 = e^0 - e^0 = 1 - 1,\text{ i.e. }y' (0) = 0\]

Hence, y = ex + e−x is the solution to the given initial value problem.