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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. ${\text{\hspace{0.17em}\hspace{0.17em}}}\frac{\mathrm{dy}}{\mathbf{d}\theta }{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{y}}}^{2}$

The solution of the given equation is $\mathbf{y}\mathbf{=}\frac{2}{\mathbf{1}\mathbf{+}{\mathbf{Ce}}^{\mathbf{2}\theta }}$.

See the step by step solution

## Step 1: Given information and simplification

Given that, $\frac{\mathrm{dy}}{\mathbf{d}\theta }\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}{\mathbf{y}}^{2}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$ $\frac{\mathrm{dy}}{\mathbf{d}\theta }\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}{\mathbf{y}}^{2}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Since, the given equation is the Bernoulli equation with n = 2, $\mathbf{P}\left(\theta \right)\mathbf{=}\mathbf{2}$ and $\mathbf{Q}\left(\theta \right)\mathbf{=}\mathbf{1}$ .

Now divide y2 on both sides of the equation (1).

${\mathbf{y}}^{-2}\frac{\mathrm{dy}}{\mathbf{d}\theta }\mathbf{+}\mathbf{2}{\mathbf{y}}^{-1}\mathbf{=}\mathbf{1}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Let us take $\mathbf{u}\mathbf{=}{\mathbf{y}}^{-1}$ and $\mathbf{-}\frac{\mathrm{du}}{\mathbf{d}\theta }\mathbf{=}{\mathbf{y}}^{-2}\frac{\mathrm{dy}}{\mathbf{d}\theta }$. Then,

$\begin{array}{c}\mathbf{-}\frac{\mathrm{du}}{\mathbf{d}\theta }\mathbf{+}\mathbf{2}\mathbf{u}\mathbf{=}\mathbf{1}\\ \frac{\mathrm{du}}{\mathbf{d}\theta }\mathbf{-}\mathbf{2}\mathbf{u}\mathbf{=}\mathbf{-}\mathbf{1}\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)\end{array}$

Let $\mathbf{P}\left(\theta \right)\mathbf{=}\mathbf{-}\mathbf{2}$

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

$\begin{array}{c}\mu \left(\theta \right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(\theta \right)\mathbf{d}\theta }\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\int 2\mathbf{d}\theta }\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}2\theta }\end{array}$

Multiply ${\mathbf{e}}^{\mathbf{-}2\theta }$ in equation (3) on both sides.

$\begin{array}{c}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\theta }\frac{\mathrm{du}}{\mathbf{d}\theta }\mathbf{-}\mathbf{2}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\theta }\mathbf{u}\mathbf{=}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\theta }\\ \frac{\mathrm{du}}{\mathbf{d}\theta }\left[{\mathbf{e}}^{\mathbf{-}\mathbf{2}\theta }\mathbf{u}\right]\mathbf{=}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\theta }\end{array}$

## Step 2: integration method

Now integrate the equation on both sides.

$\begin{array}{c}\int \frac{\mathrm{du}}{\mathbf{d}\theta }\left[{\mathbf{e}}^{\mathbf{-}\mathbf{2}�}\mathbf{u}\right]\mathbf{d}\theta \mathbf{=}\mathbf{-}\int {\mathbf{e}}^{\mathbf{-}\mathbf{2}\theta }\mathbf{d}\theta \\ {\mathbf{e}}^{\mathbf{-}\mathbf{2}\theta }\mathbf{u}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{2}\theta }}{2}\mathbf{+}{\mathbf{C}}_{1}\\ \mathbf{u}\mathbf{=}\frac{1}{2}\mathbf{+}{\mathbf{C}}_{1}{\mathbf{e}}^{\mathbf{2}\theta }\\ {\mathbf{y}}^{-1}\mathbf{=}\frac{\mathbf{1}\mathbf{+}{\mathbf{Ce}}^{\mathbf{2}\theta }}{2}\\ \mathbf{y}\mathbf{=}\frac{2}{\mathbf{1}\mathbf{+}{\mathbf{Ce}}^{\mathbf{2}\theta }}\end{array}$

So, the solution is ${\mathbf{y}}{\mathbf{=}}\frac{2}{\mathbf{1}\mathbf{+}{\mathbf{Ce}}^{\mathbf{2}\theta }}$