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Question
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x^3 \frac{d^2 y}{d x^2} = 1\]
|
\[y = ax + b + \frac{1}{2x}\]
|
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Solution
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.
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