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Found in: Page 79

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Question: In Problems 1-30, solve the equation.-${\text{\hspace{0.17em}}}\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{-}}{\mathbf{4}}{\mathbf{y}}{\mathbf{=}}{\mathbf{32}}{{\mathbf{x}}}^{2}$

The solution of the given equation is $\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{8}{\mathbf{x}}^{2}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{+}{\mathbf{Ce}}^{4x}$

See the step by step solution

## Step 1: Given information and simplification

Given that, $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{32}{\mathbf{x}}^{2}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Let P( x ) = -4.

Find the value of $\mu \left(x\right)$.

$\begin{array}{c}\mu \left(x\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(x\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\int \mathbf{4}\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{-4x}\end{array}$

Multiply e-4x in equation (1) on both sides.

$\begin{array}{c}{\mathbf{e}}^{-4x}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{-}\mathbf{4}{\mathbf{e}}^{-4x}\mathbf{y}\mathbf{=}\mathbf{32}{\mathbf{e}}^{-4x}{\mathbf{x}}^{2}\\ \frac{d}{\mathrm{dx}}\left[{\mathbf{e}}^{-4x}\mathbf{y}\right]\mathbf{=}\mathbf{32}{\mathbf{e}}^{-4x}{\mathbf{x}}^{2}\end{array}$

Now integrate the equation on both sides.

$\begin{array}{c}\int \frac{d}{\mathrm{dx}}\left[{\mathbf{e}}^{-4x}\mathbf{y}\right]\mathbf{dx}\mathbf{=}\mathbf{32}\int {\mathbf{e}}^{-4x}{\mathbf{x}}^{2}\mathbf{dx}\\ {\mathbf{e}}^{-4x}\mathbf{y}\mathbf{=}\mathbf{32}\int {\mathbf{e}}^{-4x}{\mathbf{x}}^{2}\mathbf{dx}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\end{array}$

## Step 2: Evaluation method

Find the value of $\mathbf{32}\int {\mathbf{e}}^{-4x}{\mathbf{x}}^{2}\mathbf{dx}$ separately.

Let us take $\mathbf{u}\mathbf{=}{\mathbf{x}}^{2}\mathbf{,}\mathbf{dv}\mathbf{=}{\mathbf{e}}^{-4x}\mathbf{dx}$.

$\mathbf{du}\mathbf{=}\mathbf{2}\mathbf{x}\text{ }\mathbf{dx}\mathbf{,}\mathbf{v}\mathbf{=}\mathbf{-}\frac{{\mathbf{e}}^{-4x}}{4}$.

Use the integration by parts formula.

$\begin{array}{l}\mathbf{32}\int {\mathbf{e}}^{-4x}{\mathbf{x}}^{2}\mathbf{dx}\mathbf{=}\mathbf{32}\left[\frac{\mathbf{-}{\mathbf{x}}^{2}{\mathbf{e}}^{-4x}}{4}\mathbf{+}\int \frac{{\mathbf{xe}}^{-4x}}{2}\mathbf{dx}\right]\\ \mathbf{32}\int {\mathbf{e}}^{-4x}{\mathbf{x}}^{2}\mathbf{dx}\mathbf{=}\mathbf{32}\left[\frac{\mathbf{-}{\mathbf{x}}^{2}{\mathbf{e}}^{-4x}}{4}\mathbf{+}\frac{1}{2}\int {\mathbf{xe}}^{-4x}\mathbf{dx}\right]\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)\end{array}$

Again integrate $\int {\mathbf{xe}}^{-4x}\mathbf{dx}$ separately.

$\begin{array}{l}\mathbf{u}\mathbf{=}\mathbf{x}\mathbf{,}\mathbf{dv}\mathbf{=}{\mathbf{e}}^{-4x}\\ \mathbf{du}\mathbf{=}\mathbf{dx}\mathbf{,}\mathbf{v}\mathbf{=}\mathbf{-}\frac{{\mathbf{e}}^{-4x}}{4}\end{array}$

Use the integration by parts method.

$\begin{array}{c}\int {\mathbf{xe}}^{-4x}\mathbf{dx}\mathbf{=}\mathbf{-}\frac{{\mathbf{xe}}^{-4x}}{4}\mathbf{+}\frac{1}{4}\int {\mathbf{e}}^{-4x}\mathbf{dx}\\ \mathbf{=}\mathbf{-}\mathbf{x}\frac{{\mathbf{e}}^{-4x}}{4}\mathbf{+}\frac{1}{4}\left[\mathbf{-}\frac{{\mathbf{e}}^{-4x}}{4}\right]\mathbf{+}{\mathbf{C}}_{1}\end{array}$

Substitute in equation (3)

$\begin{array}{c}\mathbf{32}\int {\mathbf{e}}^{-4x}{\mathbf{x}}^{2}\mathbf{dx}\mathbf{=}\mathbf{32}\left[\frac{\mathbf{-}{\mathbf{x}}^{2}{\mathbf{e}}^{-4x}}{4}\mathbf{+}\frac{1}{2}\left[\mathbf{-}\mathbf{x}\frac{{\mathbf{e}}^{-4x}}{4}\mathbf{+}\frac{1}{4}\left[\mathbf{-}\frac{{\mathbf{e}}^{-4x}}{4}\right]\mathbf{+}{\mathbf{C}}_{1}\right]\right]\\ \mathbf{=}\mathbf{32}\left[\mathbf{-}{\mathbf{x}}^{2}\frac{{\mathbf{e}}^{-4x}}{4}\mathbf{-}\mathbf{x}\frac{{\mathbf{e}}^{-4x}}{8}\mathbf{-}\frac{{\mathbf{e}}^{-4x}}{32}\mathbf{+}\mathbf{C}\right]\\ \mathbf{=}\mathbf{-}\mathbf{8}{\mathbf{x}}^{2}{\mathbf{e}}^{-4x}\mathbf{-}\mathbf{4}{\mathbf{xe}}^{-4x}\mathbf{-}{\mathbf{e}}^{-4x}\mathbf{+}\mathbf{C}\end{array}$

Now substitute in equation (2)

$\begin{array}{c}{\mathbf{e}}^{-4x}\mathbf{y}\mathbf{=}\mathbf{32}\int {\mathbf{e}}^{-4x}{\mathbf{x}}^{2}\mathbf{dx}\\ {\mathbf{e}}^{-4x}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{8}{\mathbf{x}}^{2}{\mathbf{e}}^{-4x}\mathbf{-}\mathbf{4}{\mathbf{xe}}^{-4x}\mathbf{-}{\mathbf{e}}^{-4x}\mathbf{+}\mathbf{C}\\ \mathbf{y}\mathbf{=}\frac{\mathbf{-}\mathbf{8}{\mathbf{x}}^{2}{\mathbf{e}}^{-4x}\mathbf{-}\mathbf{4}{\mathbf{xe}}^{-4x}\mathbf{-}{\mathbf{e}}^{-4x}\mathbf{+}\mathbf{C}}{{\mathbf{e}}^{-4x}}\\ \mathbf{y}\mathbf{=}\mathbf{-}\mathbf{8}{\mathbf{x}}^{2}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{+}{\mathbf{Ce}}^{4x}\end{array}$

So, the solution is data-custom-editor="chemistry" ${\mathbf{y}}{\mathbf{=}}{\mathbf{-}}{\mathbf{8}}{{\mathbf{x}}}^{2}{\mathbf{-}}{\mathbf{4}}{\mathbf{x}}{\mathbf{-}}{\mathbf{1}}{\mathbf{+}}{{\mathbf{Ce}}}^{4x}$

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