Q36E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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) $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}$ , x > 0 by completing the following steps:

(b) Find a nontrivial solution  ${\mathbf{y}}_{\mathbf{h}}$ to the separable equation (22) $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{0}$  , $\mathbf{x}\mathbf{>}\mathbf{0}$ .

(c) Assuming (21) has a solution of the form${\mathbf{y}}_{\mathbf{p}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$  , substitute this into equation (21), and simplify to obtain  $\mathbf{v}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}$. 

d) Now integrate to get $\mathbf{v}\left(\mathbf{x}\right)\mathbf{ }$ 

(e) Verify that $\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{C}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$  is a general solution to (21).

 

Step-by-Step Solution

Answer

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

Step 1(a): Show that the general solution to (20) dy dx + P ( x ) y = Q ( x ) has the form y ( x ) = C y h ( x ) + y p ( 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 

 $$\begin{gathered} \frac{\mathrm{d}\mu }{\mathrm{dx}}=\mathrm{P}\left(\mathrm{x}\right)\mu \left(\mathrm{x}\right) \\ \frac{{\mathrm{dy}}_{\mathrm{h}}}{\mathrm{dx}}=-\mu {\left(\mathrm{x}\right)}^{-1}\mathrm{P}\left(\mathrm{x}\right) \\ \frac{{\mathrm{dy}}_{\mathrm{h}}}{\mathrm{dx}}=-\mathrm{y}\left(\mathrm{x}\right)\mathrm{P}\left(\mathrm{x}\right) \end{gathered}$$

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

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

 

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

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 

 $$\begin{gathered} \int \mathbf{-}\frac{\mathbf{1}}{\mathbf{3}\mathbf{y}}\mathbf{dy}\mathbf{=}\int \frac{\mathbf{dx}}{\mathbf{x}} \\ \mathbf{ln} \left|\mathbf{y}\right|\mathbf{=}\mathbf{ln} \left| {\mathbf{x}}^{\mathbf{-}\mathbf{3}}\right|\mathbf{-}\mathbf{C} \end{gathered}$$

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

 (k is arbitrary constants)

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

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

 $$\begin{gathered} \frac{{\mathbf{dy}}_{\mathbf{p}}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}{\mathbf{y}}_{\mathbf{p}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}} \\ \mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{{\mathbf{dy}}_{\mathbf{h}}}{\mathbf{dx}}\mathbf{+}{\mathbf{y}}_{\mathbf{h}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}{\mathbf{y}}_{\mathbf{h}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}} \\ \mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\left[\frac{{\mathbf{dy}}_{\mathbf{h}}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}{\mathbf{y}}_{\mathbf{h}}\right]\mathbf{+}{\mathbf{y}}_{\mathbf{h}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}} \end{gathered}$$

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

 

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

Integrate the value 

     $$\begin{gathered} \frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}{\mathbf{x}}^{\mathbf{6}} \\ \mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{6}}}{\mathbf{6}} \end{gathered}$$   

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

Step 5(e): Verify that y ( x ) = C y h ( x ) + v ( x ) y h ( x ) is a general solution to (21)

Since 

$$\begin{gathered} \mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{C}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{y}}_{\mathbf{h}}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{C}{\mathbf{x}}^{\mathbf{-}\mathbf{3}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{6}}}{\mathbf{6}} \end{gathered}$$

Calculate

 $$\begin{gathered} \frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}\mathbf{y} \\ \left({\mathbf{Cx}}^{\mathbf{-}\mathbf{3}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{6}}}{\mathbf{6}}\right)\left(\mathbf{-}\mathbf{3}{\mathbf{Cx}}^{\mathbf{-}\mathbf{4}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}{\mathbf{Cx}}^{\mathbf{-}\mathbf{3}}\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}} \end{gathered}$$

 Therefore, this verifies the result.