Q21 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine which values of m the function $\mathit{\varphi }\left(x\right)\mathbf{=}{\mathbf{x}}^{\mathbf{m}}$  is a solution to the given equation.

(a) $\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{11}\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}$

(b) ${\mathbf{x}}^{\mathbf{2}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{-}\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

1. $\mathrm{m}=\frac{1}{3}, -3$
   
2. $\mathrm{m}=1\pm \sqrt{6}$
   

Step 1 (a): Taking the given function as y

First of all, we will take $\varphi \left(\mathrm{x}\right)=\mathrm{y}$

Step 2: Differentiating the given function

Differentiating concerning x,

 $\varphi \text{'}\left(\mathrm{x}\right)=\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m} {\mathrm{x}}^{\mathrm{m}-1}$

 

Again, differentiating concerning x,

$\varphi \text{'}\text{'}\left(\mathrm{x}\right)=\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\mathrm{m}\left(\mathrm{m}-1\right){\mathrm{x}}^{\mathrm{m}-2}$

Step 3: Substituting the values from step 2 in the given differential equation

$\begin{array}{l}3{\mathrm{x}}^2\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+11\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-3\mathrm{y}=0\\ 3{\mathrm{x}}^2\left[\mathrm{m}\left(\mathrm{m}-1\right){\mathrm{x}}^{\mathrm{m}-2}\right]+11\left(\mathrm{x}\right){\mathrm{mx}}^{\mathrm{m}-1}-3{\mathrm{x}}^{\mathrm{m}}=0\\ 3{\mathrm{m}}^2{\mathrm{x}}^{\mathrm{m}}-3{\mathrm{mx}}^{\mathrm{m}}+11{\mathrm{mx}}^{\mathrm{m}}-3{\mathrm{x}}^{\mathrm{m}}=0\\ 3{\mathrm{m}}^2+8\mathrm{m}-3=0\\ \left(\mathrm{m}-\frac{1}{3}\right)\left(\mathrm{m}+3\right)=0\\ \mathrm{m}=\frac{1}{3},-3\end{array}$

Hence, the values of m are $\frac{\mathbf{1}}{\mathbf{3}}$ and -3.

Step 4(b): Taking the given function as y

First of all, we will take $\varphi \left(\mathrm{x}\right)=\mathrm{y}$.

Step 5: Differentiating the given function

Differentiating concerning x,

$\varphi \text{'}\left(\mathrm{x}\right)=\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m} {\mathrm{x}}^{\mathrm{m}-1}$

Again, differentiating concerning x,

$\varphi \text{'}\text{'}\left(\mathrm{x}\right)=\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\mathrm{m}\left(\mathrm{m}-1\right){\mathrm{x}}^{\mathrm{m}-2}$

Step 6: Substituting the values from step 2 in the given differential equation.

$\begin{array}{l}{\mathrm{x}}^2\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-5\mathrm{y}=0\\ {\mathrm{x}}^2\mathrm{m}\left(\mathrm{m}-1\right){\mathrm{x}}^{\mathrm{m}-2}-{\mathrm{xmx}}^{\mathrm{m}-1}-5{\mathrm{x}}^{\mathrm{m}}=0\\ {\mathrm{m}}^2{\mathrm{x}}^{\mathrm{m}}-{\mathrm{mx}}^{\mathrm{m}}-{\mathrm{mx}}^{\mathrm{m}}-5{\mathrm{x}}^{\mathrm{m}}=0\\ {\mathrm{m}}^2{\mathrm{x}}^{\mathrm{m}}-2{\mathrm{mx}}^{\mathrm{m}}-5{\mathrm{x}}^{\mathrm{m}}=0\\ \left({\mathrm{m}}^2-2\mathrm{m}-5\right){\mathrm{x}}^{\mathrm{m}}=0\\ {\mathrm{m}}^2-2\mathrm{m}-5=0\\ \mathrm{m}=1\pm \sqrt{6}\end{array}$

Hence, the values of m are $\left(1+\sqrt{6}\right)$ and $\left(1-\sqrt{6}\right)$.