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Q2.6-2E

Expert-verifiedFound in: Page 76

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**In problems 1 - 8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form ${{\mathbf{Y}}}^{{\mathbf{\text{'}}}}{\mathbf{=}}{\mathbf{G}}{\left(\mathrm{ax}+\mathrm{by}\right)}$. ${{\left(y-4x-1\right)}}^{{\mathbf{2}}}{\mathbf{dx}}{\mathbf{-}}{\mathbf{dy}}{\mathbf{=}}{\mathbf{0}}$.**

The given equation is the form of linear coefficient.

- Homogeneous equation

If the right-hand side of the equation $\frac{dy}{dx}=f\left(x,y\right)$can be expressed as a function of the ratio $\frac{y}{x}$ alone, then we say the equation is homogeneous.

Equations of the form $\frac{dy}{dx}=G\left(ax+by\right)$

When the right-hand side of the equation $\frac{dy}{dx}=f\left(x,y\right)$can be expressed as a function of the combination $ax+by$, where a and b are constants, that is, $\frac{dy}{dx}=G\left(ax+by\right)$then the substitution $z=ax+by$ transforms the equation into a separable one.

- Bernoulli’s equation

A first-order equation that can be written in the form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right){y}^{n}$, where P(x) and Q(x) are continuous on an interval (a,b) and n is a real number, is called a Bernoulli equation.

- Equation of Linear coefficients

We have used various substitutions for y to transform the original equation into a new equation that we could solve. In some cases, we must transform both x and y into new variables, say u and v. This is the situation for equations with linear coefficients-that is, equations of the form

$\left({a}_{1}x+{b}_{1}y+{c}_{1}\right)dx+\left({a}_{2}x+{b}_{2}y+{c}_{2}\right)dy=0$

Given, ${\left(y-4x-1\right)}^{2}dx-dy=0$

By Evaluating,

role="math" localid="1655099619700" $\begin{array}{rcl}{\left(y-4x-1\right)}^{2}dx-dy& =& 0\\ \frac{dy}{dx}& =& {\left(y-4x-1\right)}^{2}\end{array}$

Let $u=y-4x$. Then, differentiate it to find the value of $\frac{du}{dx}$

$\frac{du}{dx}=\frac{dy}{dx}-4\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=\frac{du}{dx}+4$

Now,

$\begin{array}{rcl}\frac{du}{dx}+4& =& {\left(u-1\right)}^{2}\\ \frac{du}{dx}& =& \left({u}^{2}-2u+1\right)-4\\ & =& {u}^{2}-2u-3\\ \frac{du}{{u}^{2}-2u-3}& =& dx\end{array}$.....................(1)

Integration equation (1),

$\int \frac{du}{{u}^{2}-2u-3}=\int dx\phantom{\rule{0ex}{0ex}}In\left[{\left[\frac{3-u}{u+1}\right]}^{\frac{1}{4}}\right]=x+C$

role="math" localid="1655100433864" $\begin{array}{rcl}\frac{3-u}{u+1}& =& C{e}^{4x}\\ -1+\frac{4}{u+1}& =& C{e}^{4x}\\ \frac{4}{u+1}& =& C{e}^{4x}+1\\ u+1& =& \frac{4}{C{e}^{4x}+1}\\ u& =& \frac{4}{C{e}^{4x}+1}-1\end{array}$

substitute $u=y-4x$

role="math" localid="1655100444784" $\begin{array}{rcl}y-4x& =& \frac{4}{C{e}^{4x}+1}-1\\ y& =& \frac{4}{C{e}^{4x}+1}+4x-1\end{array}$

It seems that the given equation is linear coefficient.

**Therefore, the given equation is the form of linear coefficient.**

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