Question

Solve the second order equation

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

Accepted Answer

We have, $e^{- x} \frac{d^{2} y}{d x^{2}} = x$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = x e^{x}$

On integrating both sides, we get:

$\Rightarrow \int \frac{d^{2} y}{d x^{2}} d x = \int x e^{x} d x$

$\Rightarrow \frac{d y}{d x} = x \int e^{x} d x - \int [\frac{d x}{d x} \int e^{x} d x] d x$ [Applying Integration By parts by taking $x$ as first function and $e^{x}$ as second function.]

$\Rightarrow \frac{d y}{d x} = x e^{x} - \int [e^{x}] d x$

$\Rightarrow \frac{d y}{d x} = x e^{x} - e^{x} + A$

$\Rightarrow \frac{d y}{d x} = e^{x} (x - 1) + A$

On integrating both sides, we get:

$\int 1 \cdot d y = \int (e^{x} (x - 1) + A) d x$

$\Rightarrow y = \int e^{x} (x - 1) d x + \int A d x$

$\Rightarrow y = (x - 1) \int e^{x} d x - \int [\frac{d (x - 1)}{d x} \int e^{x} d x] d x + A x + B$ [Applying Integration By parts by taking $(x - 1)$ as first function and $e^{x}$ as second function.]

$\Rightarrow y = (x - 1) e^{x} - \int [e^{x}] d x + A x + B$

$\Rightarrow y = (x - 1) e^{x} - e^{x} + A x + B$

$\Rightarrow y = x e^{x} - e^{x} - e^{x} + A x + B$

$\Rightarrow y = x e^{x} - 2 e^{x} + A x + B$

Odisha Board Solutions for Chapter: Differential Equations, Exercise 1: EXERCISE-11(a)

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Differential Equations, Exercise 1: EXERCISE-11(a)

Questions from Odisha Board Solutions for Chapter: Differential Equations, Exercise 1: EXERCISE-11(a) with Hints & Solutions

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