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.

$\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}{\mathbf{e}}^{\mathbf{x}\mathbf{/}\mathbf{2}}\mathbf{,}{\mathbf{xe}}^{\mathbf{x}\mathbf{/}\mathbf{2}}\mathbf{,}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$

Short answer

The given functions satisfy the given D.E and are linearly independently on the interval$\left(-\infty ,\infty \right)andy=c_1{e}^{\frac{1}{2}x}+c_{2}x{e}^{\frac{1}{2}x}$ .

Step-by-step solution

Verify the function for the given differential equation

The third order differential equation

$4y\text{'}\text{'}-4y\text{'}+y=0$,

With the functions$y_1=e^{\frac{1}{2}x}and{y}_2=x{e}^{\frac{1}{2}x}$ , and we have to verify the functions form a fundamental set of solutions of the given differential equation on the interval $(-\infty ,\infty )$as the following technique:

Verify the given functions satisfy the given differential equation as the following:

The first function$y_1=e^{\frac{1}{2}x}$

Differentiating the function with respect to x, then we have,

$y_1^{\text{'}}=\frac{1}{2}{e}^{\frac{1}{2}x}$

And

$y_1^{\text{'}\text{'}}=\frac{1}{4}{e}^{\frac{1}{2}x}$

Substitute the derivatives to the given differential equation. Then we have

$\begin{array}{rcl}4\times \frac{1}{4}{e}^{\frac{1}{2}x}-4\times \frac{1}{2}{e}^{\frac{1}{2}x}+e^{\frac{1}{2}x}& =& 0\\ e^{\frac{1}{2}x}-2e^{\frac{1}{2}x}+e^{\frac{1}{2}x}& =& 0\\ 2e^{\frac{1}{2}x}-2e^{\frac{1}{2}x}& =& 0\\ 0& =& 0\end{array}$

The function $y_1=e^{\frac{1}{2}x}$ satisfies the given differential equation.

Solve the second function

The second function$y_2=x{e}^{\frac{1}{2}x}$

Differentiating the function with respect to x, then we have,

$y_1^{\text{'}}=e^{\frac{1}{2}x}+\frac{1}{2}x{e}^{\frac{1}{2}x}$

And

$\begin{array}{rcl}{y}_1^{\text{'}\text{'}}& =& \frac{1}{2}{e}^{\frac{1}{2}x}+\frac{1}{2}{e}^{\frac{1}{2}x}+\frac{1}{4}x{e}^{\frac{1}{2}x}\\ & =& e^{\frac{1}{2}x}+\frac{1}{4}x{e}^{\frac{1}{2}x}\end{array}$

Substitute the derivatives to the given differential equation. Then we have

$\begin{array}{rcl}4\times \left(e^{\frac{1}{2}x}+\frac{1}{4}x{e}^{\frac{1}{2}x}\right)-4\times \left(e^{\frac{1}{2}x}+\frac{1}{2}x{e}^{\frac{1}{2}x}\right)+x{e}^{\frac{1}{2}x}& =& 0\\ 4e^{\frac{1}{2}x}+x{e}^{\frac{1}{2}x}-4e^{\frac{1}{2}x}-2x{e}^{\frac{1}{2}x}+x{e}^{\frac{1}{2}x}& =& 0\\ 4e^{\frac{1}{2}x}-4e^{\frac{1}{2}x}-2x{e}^{\frac{1}{2}x}+2x{e}^{\frac{1}{2}x}& =& 0\\ 0& =& 0\end{array}$

The function $y_2=x{e}^{\frac{1}{2}x}:$satisfies the given differential equation.

Find the given function is linearly dependant or independent Using Wronskian:

$\begin{array}{rcl}W\left(y_1,y_2\right)& =& \left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\text{'}}& y_2^{\text{'}}\end{array}\right|\\ & =& \left|\begin{array}{cc}{e}^{\frac{1}{2}x}& x{e}^{\frac{1}{2}x}\\ \frac{1}{2}{e}^{\frac{1}{2}x}& e^{\frac{1}{2}x}+\frac{1}{2}x{e}^{\frac{1}{2}x}\end{array}\right|\end{array}$

Determinate the matrix as

$\begin{array}{rcl}W\left(x^3,x^4\right)& =& e^{\frac{1}{2}x}\times \left(e^{\frac{1}{2}x}+\frac{1}{2}x{e}^{\frac{1}{2}x}\right)-x{e}^{\frac{1}{2}x}\times \frac{1}{2}{e}^{\frac{1}{2}x}\\ & =& e^x+\frac{1}{2}x{e}^x-\frac{1}{2}x{e}^x\\ & =& e^x\end{array}$

Since the determinate of the Wronskian of the given set of functions is not equal Zero, then this set of function is linearly independent,

The general solution is$y=c_1{e}^{\frac{1}{2}x}+c_{2}x{e}^{\frac{1}{2}x}$ .