Question

Differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 0, y \left( 0 \right) = 2, y'\left( 0 \right) = 1\]

Function y = e^x + 1

Answer 1

We have,

\[y = e^x + 1...........(1)\]

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

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

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

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

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

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

It is the given differential equation.

\[y = e^x + 1\] satisfies the given differential equation; hence, it is a solution.

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

And, when \[x = 0, y' = e^0 = 1,\text{ i.e. }y'(0) = 1\]

Hence, \[y = e^x + 1\] is the solution to the given initial value problem.