Question

For the following differential equation verify that the accompanying function is a solution:

Differential equation	Function
\[x^3 \frac{d^2 y}{d x^2} = 1\]	\[y = ax + b + \frac{1}{2x}\]

Answer 1

We have,

\[y = ax + b + \frac{1}{2x} . . . . . \left( 1 \right)\]

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

\[\frac{dy}{dx} = a - \frac{1}{2 x^2} . . . . . \left( 2 \right)\]

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

\[ \Rightarrow \frac{d^2 y}{d x^2} = \left( - \frac{1}{2} \right) \times \left( \frac{- 2}{x^3} \right)\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{1}{x^3}\]

\[ \Rightarrow x^3 \frac{d^2 y}{d x^2} = 1\]

Hence, the given function is the solution to the given differential equation.