Question

In Problems 23 - 30 verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. Form the general solution.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{12}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}{\mathbf{e}}^{\mathbf{-}\mathbf{3}\mathbf{x}}\mathbf{,}{\mathbf{e}}^{\mathbf{4}\mathbf{x}}\mathbf{,}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$

Short answer

The solution of the given function to form the general solution is $y=c_1{e}^{-3x}+c_2{e}^{4x}$.

Step-by-step solution

Verify the function for the given differential equation

For the first function$y_1\left(x\right)=e^{-3x}$ , we have,

$y\text{'}=-3e^{-3x}\mathrm{and}y\text{'}\text{'}=9e^{-3x}$

Substituting the derivatives in the DE gives,

$\begin{array}{rcl}y\text{'}\text{'}-y\text{'}-12y& =& 0\\ 9e^{-3x}-\left(-3e^{-3x}\right)-12\left(e^{-3x}\right)& =& 0\\ 9e^{-3x}+3e^{-3x}-12e^{-3x}& =& 0\\ 0& =& 0\end{array}$

Similarly for the second function$y_2\left(x\right)=e^{4x}$ , we have

$y\text{'}=4e^{4x}\mathrm{and}y\text{'}\text{'}=16e^{4x}$

Substituting the derivatives in the DE gives,

$\begin{array}{rcl}y\text{'}\text{'}-y\text{'}-12y& =& 0\\ 16e^{4x}-4e^{4x}-12\left(e^{4x}\right)& =& 0\\ 12e^{4x}-12e^{4x}& =& 0\\ 0& =& 0\end{array}$

Both the function satisfies the DE, the functions$y=e^{-3x}\mathrm{and}y=e^{4x}$ are the solutions of the differential equation.

Determine the set of function by using Wronskian

The coefficient functions${\mathrm{a}}_2\left(\mathrm{x}\right)=1,{\mathrm{a}}_1\left(\mathrm{x}\right)=-1\mathrm{and}{\mathrm{a}}_0\left(\mathrm{x}\right)=-12$are continuous on$(-\infty ,\infty )$and that$a_2\left(x\right)\ne 0$ for every value of x in the interval. Wronskian is

$\begin{array}{rcl}W\left(e^{-3x},e^{4x}\right)& =& \left|\begin{array}{cc}{e}^{-3x}& e^{4x}\\ -3e^{-3x}& 4e^{4x}\end{array}\right|\\ & =& e^{-3x}·4e^{4x}-e^{4x}·\left(-3e^{-3x}\right)\\ & =& 4e^x+3e^x\\ & =& 7e^x\end{array}$

Since$W\left(y_1\left(x\right),y_2\left(x\right)\right)=7e^x\ne 0$ for all x, we can conclude that$y_1\left(x\right)\mathrm{and}{y}_2\left(x\right)$are linearly independent on$(-\infty ,\infty )$ .Hence,$e^{-3x}\mathrm{and}{e}^{4x}$form a fundamental set of solutions of the differential equation on the indicated interval.

The general solution of the equation on the interval is,

$y=c_1{e}^{-3x}+c_2{e}^{4x}$

Where$c_1,c_2$ are arbitrary constants.