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### 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

# Variation of Parameters. Here is another procedure for solving linear equations that is particularly useful for higher-order linear equations. This method is called variation of parameters. It is based on the idea that just by knowing the form of the solution, we can substitute into the given equation and solve for any unknowns. Here we illustrate the method for first-order equations (see Sections 4.6 and 6.4 for the generalization to higher-order equations).(a) Show that the general solution to (20)$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ has the form$\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{C}{{\mathbf{y}}}_{{\mathbf{h}}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}{{\mathbf{y}}}_{{\mathbf{p}}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, where ${{\mathbf{y}}}_{{\mathbf{h}}}$ ( ${\mathbf{\ne }}{\mathbf{0}}$is a solution to equation (20) when ${\mathbf{Q}}{\mathbf{\left(}}{\mathbf{x}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}{\mathbf{}}$, C is a constant, and${{\mathbf{y}}}_{{\mathbf{p}}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{{\mathbf{y}}}_{{\mathbf{h}}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ for a suitable function v(x). [Hint: Show that we can take ${{\mathbf{y}}}_{{\mathbf{h}}}{\mathbf{=}}{{\mathbf{\mu }}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ and then use equation (8).] We can in fact determine the unknown function ${\mathbf{y}}_{\mathbf{h}}$by solving a separable equation. Then direct substitution of v${\mathbf{y}}_{\mathbf{h}}$ in the original equation will give a simple equation that can be solved for v.Use this procedure to find the general solution to (21) localid="1663920708127" $\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{+}}\frac{\mathbf{3}}{\mathbf{x}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{x}}}^{{\mathbf{2}}}$ , x > 0 by completing the following steps:(b) Find a nontrivial solution ${\mathbf{y}}_{\mathbf{h}}$ to the separable equation (22) localid="1663920724944" $\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{+}}\frac{\mathbf{3}}{\mathbf{x}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$ , localid="1663920736626" ${\mathbf{x}}{\mathbf{>}}{\mathbf{0}}$ .(c) Assuming (21) has a solution of the formlocalid="1663920777078" ${{\mathbf{y}}}_{{\mathbf{p}}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{{\mathbf{y}}}_{{\mathbf{h}}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ , substitute this into equation (21), and simplify to obtain localid="1663920789271" $\mathbf{v}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}$. d) Now integrate to get localid="1663920800433" ${\mathbf{v}}\left(\mathbf{x}\right){\mathbf{}}$ (e) Verify that localid="1663920811828" $\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{C}{{\mathbf{y}}}_{{\mathbf{h}}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{{\mathbf{y}}}_{{\mathbf{h}}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ is a general solution to (21).

1. Proved
2. ${\mathrm{y}}_{\mathrm{h}}=± \mathrm{k}.{\mathrm{x}}^{-3}$
3. $\frac{\mathrm{dv}}{\mathrm{dx}}={\mathrm{x}}^{6}$
4. $\mathrm{v}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^{6}}{6}$
5. Proved
See the step by step solution

## Step 1(a): Show that the general solution to (20)dydx+P(x)y=Q(x)   has the formy(x)=Cyh(x)+yp(x)

Here $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ and ${\mathbf{y}}_{\mathbf{h}}\mathbf{=}{\mu }^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

So the equation is

$\frac{{\mathbf{dy}}_{\mathbf{h}}}{\mathbf{dx}}\mathbf{=}\mathbf{-}\mu {\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{d}\mu }{\mathbf{dx}}$

For

$\frac{\mathrm{d}\mu }{\mathrm{dx}}=\mathrm{P}\left(\mathrm{x}\right)\mu \left(\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{dy}}_{\mathrm{h}}}{\mathrm{dx}}=-\mu {\left(\mathrm{x}\right)}^{-1}\mathrm{P}\left(\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{dy}}_{\mathrm{h}}}{\mathrm{dx}}=-\mathrm{y}\left(\mathrm{x}\right)\mathrm{P}\left(\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}$

Rearranging them $\frac{{\mathrm{dy}}_{\mathrm{h}}}{\mathrm{dx}}+\mathrm{y}\left(\mathrm{x}\right)\mathrm{P}\left(\mathrm{x}\right)=0$

The solution is

localid="1663919860296" width="275" style="max-width: none;" $\mathrm{y}\left(\mathrm{x}\right)=\frac{1}{\mu \left(\mathrm{x}\right)}\left[\int \mu \left(\mathrm{x}\right)\mathrm{Q}\left(\mathrm{x}\right)\mathrm{dx}+\mathrm{C}\right]\phantom{\rule{0ex}{0ex}}{\mathrm{Cy}}_{\mathrm{h}}\left(\mathrm{x}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\frac{1}{\mu \left(\mathrm{x}\right)}\left[\int \mu \left(\mathrm{x}\right)\mathrm{Q}\left(\mathrm{x}\right)\mathrm{dx}+\mathrm{C}\right]\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{v}\left(\mathrm{x}\right){\mathrm{y}}_{\mathrm{h}}\left(\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{v}\left(\mathrm{x}\right)\mu {\left(\mathrm{x}\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{\mu \left(\mathrm{x}\right)}\left[\int \mu \left(\mathrm{x}\right)\mathrm{Q}\left(\mathrm{x}\right)\mathrm{dx}+\mathrm{C}\right]\phantom{\rule{0ex}{0ex}}\mathrm{v}\left(\mathrm{x}\right)=\int \mu \left(\mathrm{x}\right)\mathrm{Q}\left(\mathrm{x}\right)\mathrm{dx}\phantom{\rule{0ex}{0ex}}$

Therefore the general solution is localid="1663920004724" ${\mathbf{v}}{ }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\int }{\mu }{ }\mathbf{\left(}\mathbf{x}\mathbf{\right)}{ }{\mathbf{Q}}{ }\mathbf{\left(}\mathbf{x}\mathbf{\right)}{ }{\mathbf{dx}}$

## Step 3(b): Find the nontrivial solution

Here $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{0}$

Rearranging them

$\int \mathbf{-}\frac{\mathbf{1}}{\mathbf{3}\mathbf{y}}\mathbf{dy}\mathbf{=}\int \frac{\mathbf{dx}}{\mathbf{x}}\phantom{\rule{0ex}{0ex}}\mathbf{ln} \left|\mathbf{y}\right|\mathbf{=}\mathbf{ln} \left| {\mathbf{x}}^{\mathbf{-}\mathbf{3}}\right|\mathbf{-}\mathbf{C}\phantom{\rule{0ex}{0ex}}$

Raising power e on both sides then${\mathbf{y}}_{\mathbf{h}}\mathbf{=}\mathbf{±} \mathbf{k}\mathbf{.}{\mathbf{x}}^{\mathbf{-}\mathbf{3}}$

(k is arbitrary constants)

Therefore the nontrivial solution is ${{\mathbf{y}}}_{{\mathbf{h}}}{\mathbf{=}}{\mathbf{±}}{ }{ }{\mathbf{k}}{\mathbf{.}}{{\mathbf{x}}}^{\mathbf{-}\mathbf{3}}$.

## Step 3(c): Determine the value of  v'(x)

Since ${\mathbf{y}}_{\mathbf{p}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{v} \mathbf{\left(}\mathbf{x}\mathbf{\right)} {\mathbf{y}}_{\mathbf{h}} \mathbf{\left(}\mathbf{x}\mathbf{\right)}$and substitute${\mathbf{y}}_{\mathbf{h}}\mathbf{=}\mathbf{v} \mathbf{\left(}\mathbf{x}\mathbf{\right)} {\mathbf{x}}^{\mathbf{-}\mathbf{3}}$ then

$\frac{{\mathbf{dy}}_{\mathbf{p}}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}{\mathbf{y}}_{\mathbf{p}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{{\mathbf{dy}}_{\mathbf{h}}}{\mathbf{dx}}\mathbf{+}{\mathbf{y}}_{\mathbf{h}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}{\mathbf{y}}_{\mathbf{h}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\left[\frac{{\mathbf{dy}}_{\mathbf{h}}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}{\mathbf{y}}_{\mathbf{h}}\right]\mathbf{+}{\mathbf{y}}_{\mathbf{h}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}$

From part (b) $\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\left[\frac{{\mathbf{dy}}_{\mathbf{h}}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}{\mathbf{y}}_{\mathbf{h}}\right]\mathbf{=}\mathbf{0}$ then

${\mathbf{y}}_{\mathbf{h}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}$

Where ${\mathbf{y}}_{\mathbf{h}}\mathbf{=}{\mathbf{x}}^{\mathbf{-}\mathbf{3}}$ then

$\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}{\mathbf{x}}^{\mathbf{6}}$

Therefore the value is role="math" localid="1663920371908" $\frac{\mathbf{dv}}{\mathbf{dx}}{\mathbf{=}}{{\mathbf{x}}}^{{\mathbf{6}}}$ .

## Step 4(d): Find the value v(x)

Integrate the value

$\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}{\mathbf{x}}^{\mathbf{6}}\phantom{\rule{0ex}{0ex}}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{6}}}{\mathbf{6}}\phantom{\rule{0ex}{0ex}}$

Therefore the value is role="math" localid="1663920445772" $\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{6}}}{\mathbf{6}}$ .

## Step 5(e): Verify that  y(x)=Cyh(x)+v(x)yh(x) is a general solution to (21)

Since

$\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{C}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{C}{\mathbf{x}}^{\mathbf{-}\mathbf{3}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{6}}}{\mathbf{6}}\phantom{\rule{0ex}{0ex}}$Calculate

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}\mathbf{y}\phantom{\rule{0ex}{0ex}}\left({\mathbf{Cx}}^{\mathbf{-}\mathbf{3}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{6}}}{\mathbf{6}}\right)\left(\mathbf{-}\mathbf{3}{\mathbf{Cx}}^{\mathbf{-}\mathbf{4}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}{\mathbf{Cx}}^{\mathbf{-}\mathbf{3}}\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}$

Therefore, this verifies the result.

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