Question

In Problems 15-28 find the general solution of the given higher order differential equation.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$

Short answer

$y=c_1+c_2{e}^{5x}+c_3{e}^{-x}$

Step-by-step solution

Define solution of higher order differential equation

General solution of a higher- order differential equation.

In order, to solve an n^th order differential equation $a_n{y}^n+a_{n-1}{y}^{n-1}+\dots a_2{y}^{\text{'}\text{'}}+a_1^{\text{'}}+a_{0}y=0$, initially solve the n^th degree polynomial equation $a_n{m}^n+a_{n-1}{m}^{n-1}+\dots \dots +a_0=0$.

If all the roots of$a_n{m}^n+a_{n-1}{m}^{n-1}+\dots \dots +a_0=0$are real and distinct, then the general solution will be$y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}+\cdots +c_n{e}^{m_{n}x}.$

If an auxiliary equation has k roots and all the roots of the auxiliary equation are equal, then the general solution will be $c_1{e}^{m_{1}x}+c_{2}x{e}^{m_{1}x}+c_3{x}^2{e}^{m_{1}x}+\dots \dots +c_k{x}^{k-1}{e}^{m_{1}x}$.

In case if higher-order differential equation has repeated conjugate complex roots. Then the general solution will be $y=c_1{e}^{m_{1}x}+c_2{e}^{-m_{1}x}+c_{3}x{e}^{m_{1}x}+c_{4}x{e}^{-m_{1}x}$, where m₁ represents an imaginary root.

Find the given differential equation

Consider the given differential equation $y\text{'}\text{'}\text{'}-4y\text{'}\text{'}-5y\text{'}=0$.

Substitute $y\text{'}\text{'}\text{'}=m^3{e}^{mx},y\text{'}\text{'}=m^2{e}^{mx}$ and $y\text{'}=m{e}^{mx}$ into $y\text{'}\text{'}\text{'}-4y\text{'}\text{'}-5y\text{'}=0$and obtain the auxiliary equation as,

$$\begin{gathered} y\text{'}\text{'}\text{'}-4y\text{'}\text{'}-5y\text{'}=0 \\ {m}^3{e}^{mx}-4m^2{e}^{mx}-5m{e}^{mx}=0 \\ {e}^{mx}\left(m^3-4m^2-5m\right)=0 \end{gathered}$$

For a real x, the value of$e^{mx}\ne 0$. Solve the auxiliary equation as follows.

$$\begin{gathered} m^3-4m^2-5m=0 \\ m\left(m^2-4m-5\right)=0 \\ m=0\text{or}{m}^2-4m-5=0 \\ m=0\text{or}(m-5)(m+1)=0 \end{gathered}$$

On further simplification,

$$\begin{gathered} m=0\text{or}(m-5)=0\text{or}(m+1)=0 \\ m=0\text{or}m=5\text{or}m=-1 \end{gathered}$$

Therefore, the roots are $m_1=0,m_2=5and{m}_3=-1$

Compute the solution of the given differential equation

That is, the roots of the auxiliary equation are real and distinct.

Then, by the above procedure, the general solution to the differential equation is,

$$\begin{gathered} y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}+\dots c_n{e}^{m_{n}x}y \\ =c_1{e}^{0·x}+c_2{e}^{5x}+c_3{e}^{-1·x}y \\ =c_1+c_2{e}^{5x}+c_3{e}^{-x} \end{gathered}$$

Thus, the general solution is $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{5}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{3}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$