Question

In Problems,1-14find the general solution of the given second-order differential equation.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

$y=c_1{e}^{2x}+c_2{e}^x$

Step-by-step solution

Step 1:DefineSecond order differential equation

For a second-order differential equation $a{y}^{\text{'}\text{'}}+b{y}^{\text{'}}+cy=0$, where aandb are constants, the auxiliary equation will be of the form $a{m}^2+bm+c=0$.

There will be three forms of the general solution corresponding to three cases thatdependon the solution of an auxiliary equation.

1.$m_1\text{and}{m}_2$are real and distinct roots that are possible when$\left(b^2-4ac>0\right)$

2.$m_1\text{and}{m}_2$are real and equal roots that are possible when$\left(b^2-4ac=0\right)$

3.$m_1\text{and}{m}_2$are conjugate complex roots$\left(b^2-4ac<0\right)$

Case I: Distinct Real Roots

The general solution of$ay\text{'}\text{'}+by\text{'}+cy=0$ on the interval$(-\infty ,\infty )$will be$y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}$.

Case II: Repeated Real Roots

The general solution will be.$y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}$

Case III: Conjugate Complex Roots

If $m_1\text{and}{m}_2$ are complex roots, then$m_1=\alpha +i\beta$and$m_2=\alpha -i\beta$, where$\alpha$and$\beta >0$are real and$i^2=-1$.

The general solution will be

$$\begin{gathered} y=c_1{e}^{\alpha x}\cos \beta x+c_2{e}^{\alpha x}\sin \beta x \\ =e^{\alpha x}\left(c_1\cos \beta x+c_2\sin \beta x\right) \end{gathered}$$

Find the given differential equation

Consider the given differential equation$y\text{'}\text{'}-3y\text{'}+2y=0$.

Consider$y=e^{mx}$as the solution of the differential equation$y\text{'}\text{'}-3y\text{'}+2y=0$. Substitute$y=e^{mx},y\text{'}=m{e}^{mx}$and$y\text{'}\text{'}=m^2{e}^{mx}$into$y\text{'}\text{'}-3y\text{'}+2y=0$to obtain the auxiliary equation,

$$\begin{gathered} m^2{e}^{mx}-3m{e}^{mx}+2e^{mx}=0 \\ {e}^{mx}\left(m^2-3m+2\right)=0 \end{gathered}$$

Solve the auxiliary equation as follows.

$$\begin{gathered} m^2-3m+2=0 \\ {m}^2-2m-m+2=0 \\ m(m-2)-1(m-2)=0 \\ (m-2)(m-1)=0 \end{gathered}$$

Solve further.

$$\begin{gathered} (m-2)=0,(m-1)=0 \\ m=2,m=1 \end{gathered}$$

The roots of the auxiliary equation are real and distinct.

Compute the solution of the given differential equation

The roots of the auxiliary equation are real and distinct.

By above procedure, compute the general solution to the differential equation.

$$\begin{gathered} y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x} \\ =c_1{e}^{2x}+c_2{e}^x \end{gathered}$$

Thus, the general solution is $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{x}}$