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Answers without the blur. Sign up and see all textbooks for free! Q 2.6-35E

Expert-verified Found in: Page 77 ### 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 33-40 Solve the equation given in Problem 3.$\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{+}}\frac{y}{x}{\mathbf{=}}{{\mathbf{x}}}^{3}{{\mathbf{y}}}^{2}$

${\mathbf{y}}{\mathbf{=}}\frac{-3}{{\mathbf{x}}^{4}\mathbf{+}\mathbf{Cx}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{\mathbf{y}}{\mathbf{\equiv }}{\mathbf{0}}$

See the step by step solution

## Step 1: Given equation solve by Bernoulli Equations.

A first-order equation that can be written in the form,

$\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{+}}{\mathbf{P}}\left(x\right){\mathbf{y}}{\mathbf{=}}{\mathbf{Q}}\left(x\right){{\mathbf{y}}}^{n}$,

where P(x) and Q(x) are continuous on an interval (a, b) and n is a real number, is called a Bernoulli equation.

## Step 2: Solve the given equation

The given equation is,

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\frac{y}{x}\mathbf{=}{\mathbf{x}}^{3}{\mathbf{y}}^{2}$

Both sides divide by y2 in the above equation,

${\mathbf{y}}^{-2}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\frac{{\mathbf{y}}^{-1}}{x}\mathbf{=}{\mathbf{x}}^{3}\text{\hspace{0.17em}}......\left(1\right)$

Let,

$\begin{array}{c}\mathbf{t}\mathbf{=}{\mathbf{y}}^{-1}\\ \frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}\mathbf{-}{\mathbf{y}}^{-2}\frac{\mathrm{dy}}{\mathrm{dx}}\end{array}$

Substitute the $\mathbf{t}\mathbf{=}{\mathbf{y}}^{-1}\text{ and }\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}\mathbf{-}{\mathbf{y}}^{-2}\frac{\mathrm{dy}}{\mathrm{dx}}$ in the equation (1),

$\begin{array}{c}\mathbf{-}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{+}\frac{t}{x}\mathbf{=}{\mathbf{x}}^{3}\\ \frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{-}\frac{t}{x}\mathbf{=}\mathbf{-}{\mathbf{x}}^{3}......\left(2\right)\end{array}$

## Step 3: Compare equation (2) with the standard form of the linear equation,

The standard form of the linear equation is,

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{P}\left(x\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(x\right)$

One has,

$\mathbf{P}\left(x\right)\mathbf{=}\mathbf{-}\frac{1}{x}$

Now find the integrating factor,

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

Both sides Multiply by $\mu \left(x\right)\mathbf{=}\frac{1}{x}$ in the equation (2),

$\frac{1}{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{-}\left(\frac{1}{x}\right)\frac{t}{x}\mathbf{=}\mathbf{-}\left(\frac{1}{x}\right){\mathbf{x}}^{3}\text{\hspace{0.17em}\hspace{0.17em}}$

Simplify the above equation,

$\begin{array}{c}\frac{1}{x}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{-}\frac{t}{{\mathbf{x}}^{2}}\mathbf{=}\mathbf{-}{\mathbf{x}}^{2}\text{\hspace{0.17em}\hspace{0.17em}}\\ \frac{d}{\mathrm{dx}}\left(\frac{t}{x}\right)\mathbf{=}\mathbf{-}{\mathbf{x}}^{2}\end{array}$

Integrate both sides and solve for the solution.

$\frac{t}{x}\mathbf{=}\mathbf{-}\int {\mathbf{x}}^{2}\mathbf{dx}\mathbf{+}\mathbf{c}$

Substitute t = y-1 and simplify the above equation,

$\begin{array}{c}\frac{{\mathbf{y}}^{-1}}{x}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^{3}}{3}\mathbf{+}\mathbf{c}\\ \frac{1}{y}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^{4}}{3}\mathbf{+}\mathbf{cx}\\ \frac{1}{y}\mathbf{=}\frac{{\mathbf{x}}^{4}+\mathbf{3}\mathbf{cx}}{-3}\end{array}$

Where, C = -3c

$\mathbf{y}\mathbf{=}\frac{-3}{{\mathbf{x}}^{4}\mathbf{+}\mathbf{Cx}}$

Hence, the solve the given equation is ${\mathbf{y}}{\mathbf{=}}\frac{-3}{{\mathbf{x}}^{4}\mathbf{+}\mathbf{Cx}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{\mathbf{y}}{\mathbf{\equiv }}{\mathbf{0}}$ ### Want to see more solutions like these? 