Q20 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

(a) $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{6}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}$

(b) $\frac{{\mathbf{d}}^{\mathbf{3}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{3}}}\mathbf{+}\mathbf{3}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{2}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

1. $\mathrm{m}=-1, -5$
   
2. $\mathrm{m}=-1, -2$
   

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 $\varphi \left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{mx}}$ with respect to x,

$\varphi \text{'}\left(\mathrm{x}\right)=\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{me}}^{\mathrm{mx}}$

Again, differentiating with respect to x,

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

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

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+6\frac{\mathrm{dy}}{\mathrm{dx}}+5\mathrm{y}=0 \\ {\mathrm{m}}^2{\mathrm{e}}^{\mathrm{mx}}+6{\mathrm{me}}^{\mathrm{mx}}+5{\mathrm{e}}^{\mathrm{mx}}=0 \\ \left({\mathrm{m}}^2+6\mathrm{m}+5\right){\mathrm{e}}^{\mathrm{mx}}=0 \\ {\mathrm{m}}^2+6\mathrm{m}+5=0 \\ \left(\mathrm{m}+1\right)\left(\mathrm{m}+5\right)=0 \\ \mathrm{m}=-1,-5 \end{gathered}$$

Hence, the values of m are -1 and -5.

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 $\varphi \left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{mx}}$ with respect to x,

$\varphi \text{'}\left(\mathrm{x}\right)=\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{me}}^{\mathrm{mx}}$

Again, differentiating with respect to x,

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

Again, differentiating with respect to x,

$\varphi \text{'}\text{'}\text{'}\left(\mathrm{x}\right)=\frac{{\mathrm{d}}^3\mathrm{y}}{{\mathrm{dx}}^3}={\mathrm{m}}^3{\mathrm{e}}^{\mathrm{mx}}$

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

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

Hence, the values of m are -1 and -2.