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Expert-verified 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 31-40, solve the initial value problem.$\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{-}}\frac{2y}{x}{\mathbf{=}}{{\mathbf{x}}}^{2}{\mathbf{cosx}}{\mathbf{,}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}\left(\pi \right){\mathbf{=}}{\mathbf{2}}$

The solution of the given equation is $\mathbf{y}\mathbf{=}{\mathbf{x}}^{2}\mathbf{sin}\text{ }\mathbf{x}\mathbf{+}\frac{\mathbf{2}{\mathbf{x}}^{2}}{{\pi }^{2}}$.

See the step by step solution

## Step 1: Given information and simplification

Given that, $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{-}\frac{2y}{x}\mathbf{=}{\mathbf{x}}^{2}\mathbf{cos}\text{ }\mathbf{x}\mathbf{,}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathbf{y}\left(\pi \right)\mathbf{=}\mathbf{2}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Let $\mathbf{P}\left(x\right)\mathbf{=}\mathbf{-}\frac{2}{x}$.

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

$\begin{array}{c}\mu \left(x\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(x\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\int \frac{2}{x}\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ln}\left|x\right|}\\ \mathbf{=}\frac{1}{{\mathbf{x}}^{2}}\end{array}$

Multiply $\frac{1}{{\mathbf{x}}^{2}}$ in equation (1) on both sides.

$\begin{array}{c}\frac{1}{{\mathbf{x}}^{2}}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{-}\frac{2y}{{\mathbf{x}}^{3}}\mathbf{=}\mathbf{cos}\text{ }\mathbf{x}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{1}{{\mathbf{x}}^{2}}\mathbf{y}\right]\mathbf{=}\mathbf{cos}\text{ }\mathbf{x}\end{array}$

## Step 2: integration method

Now integrate the equation on both sides.

${\begin{array}{c}\int \frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{1}{{\mathbf{x}}^{2}}\mathbf{y}\right]\mathbf{dx}\mathbf{=}\int \mathbf{cos}\text{ }\mathbf{x}\text{ }\mathbf{dx}\\ \frac{1}{{\mathbf{x}}^{2}}\mathbf{y}\mathbf{=}\mathbf{sin}\text{ }\mathbf{x}\mathbf{+}{\mathbf{C}}_{1}\\ \mathbf{y}\mathbf{=}{\mathbf{x}}^{2}\mathbf{sin}\text{ }\mathbf{x}\mathbf{+}{\mathbf{x}}^{2}\mathbf{C}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\end{array}}_{}$

So, the solution is found.

## Step 3: Find the initial value

Given that, $\mathbf{y}\left(\pi \right)\mathbf{=}\mathbf{2}$.

Then, $\mathbf{x}\mathbf{=}\pi$ and y = 2.

Substitute the value in equation (2) to get the value of C.

$\begin{array}{c}\mathbf{y}\mathbf{=}{\mathbf{x}}^{2}\mathbf{sin}\text{ }\mathbf{x}\mathbf{+}{\mathbf{x}}^{2}\mathbf{C}\\ \mathbf{2}\mathbf{=}{\pi }^{2}\mathbf{sin}\left(\pi \right)\mathbf{+}{\pi }^{2}\mathbf{C}\\ \frac{2}{{\pi }^{2}}\mathbf{=}\mathbf{C}\end{array}$

Substitute the value of C in equation (2).

$\mathbf{y}\mathbf{=}{\mathbf{x}}^{2}\mathbf{sin}\text{ }\mathbf{x}\mathbf{+}\frac{\mathbf{2}{\mathbf{x}}^{2}}{{\pi }^{2}}$

So, the solution is ${\mathbf{y}}{\mathbf{=}}{{\mathbf{x}}}^{2}{\mathbf{sin}}{\text{ }}{\mathbf{x}}{\mathbf{+}}\frac{\mathbf{2}{\mathbf{x}}^{2}}{{\pi }^{2}}$ ### Want to see more solutions like these? 