Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{10}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

$y=c_1{e}^{5x}+c_{2}x{e}^{5x}$

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 that depend on 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

.

Case II: Repeated Real Roots

The general solution will belocalid="1663826473701" $y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}$.

Case III: Conjugate Complex Roots

Iflocalid="1663826480041" $m_1\text{and}{m}_2$are complex roots, then $m_1=\alpha +i\beta$and $m_2=\alpha -i\beta$, where and localid="1663826144597" $\beta >0$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$y\text{'}\text{'}-10y\text{'}+25y=0$.

Consider$y=e^{mx}$ is the solution of the differential equation$y\text{'}\text{'}-10y\text{'}+25y=0$. Substitute$y=e^{mx},y\text{'}=m{e}^{mx}$and$y\text{'}\text{'}=m^2{e}^{mx}$into$y\text{'}\text{'}-10y\text{'}+25y=0$to obtain the auxiliary equation,

$$\begin{gathered} m^2{e}^{mx}-10m{e}^{mx}+25e^{mx}=0 \\ {e}^{mx}\left(m^2-10m+25\right)=0 \end{gathered}$$

Solve the auxiliary equation as follows.

$$\begin{gathered} m^2-10m+25=0 \\ {m}^2-5m-5m+25=0 \\ m(m-5)-5(m-5)=0 \\ (m-5)(m-5)=0 \end{gathered}$$

Solve further.

$$\begin{gathered} (m-5)=0,(m-5)=0 \\ m=5,5 \end{gathered}$$

Compute the solution of the given differential equation

The roots of the auxiliary equation are real and equal.

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

$$\begin{gathered} y=c_1{e}^{m_{1}x}+c_{2}x{e}^{m_{1}x} \\ =c_1{e}^{5x}+c_{2}x{e}^{5x} \end{gathered}$$

Thus, the general solutionis $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{5}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{x}{\mathit{e}}^{\mathbf{5}\mathbf{x}}$.