Question

Without the aid of the Wronskian, determine whether the given set of functions is linearly independent or linearly dependent on the indicated interval.

$$\begin{gathered} f_1\left(x\right)=x{e}^{x+1},f_2\left(x\right)=\left(4x-5\right)e^x \\ {f}_3\left(x\right)=x{e}^x,\left(-\infty ,\infty \right) \end{gathered}$$

Short answer

The set of functions$$\begin{gathered} f_1\left(x\right)=x{e}^{x+1},f_2\left(x\right)=\left(4x-5\right)e^x \\ {f}_3\left(x\right)=x{e}^x,\left(-\infty ,\infty \right) \end{gathered}$$ is linearly dependent

Step-by-step solution

Define

We can find the Wronskian of the functions with the formula as:

$W\left(f_1\left(x\right),f_2\left(x\right),f_3\left(x\right)\right)=\left|\begin{array}{ccc}f1\left(x\right)& f2\left(x\right)& f3\left(x\right)\\ f_1^{\text{'}}\left(x\right)& f_2^{\text{'}}\left(x\right)& f_3^{\text{'}}\left(x\right)\\ f_1^{"}\left(x\right)& f_2^{"}\left(x\right)& f_3^{"}\left(x\right)\end{array}\right|$

Solve the equations.

We have the functions $f_1\left(x\right)=x{e}^{x+1},f_2\left(x\right)=\left(4x-5\right)e^x$and $f_3\left(x\right)=x{e}^x$on the interval $\left(-\infty ,\infty \right)$, and we can obtain its Wronskian as

$W\left(f_1\left(x\right),f_2\left(x\right),f_3\left(x\right)\right)=\left|\begin{array}{ccc}f1\left(x\right)& f2\left(x\right)& f3\left(x\right)\\ f_1^{\text{'}}\left(x\right)& f_2^{\text{'}}\left(x\right)& f_3^{\text{'}}\left(x\right)\\ f_1^{"}\left(x\right)& f_2^{"}\left(x\right)& f_3^{"}\left(x\right)\end{array}\right|$

$=\left|\begin{array}{ccc}x{e}^{x+1}& \left(4x-5\right)e^x& x{e}^x\\ x{e}^{x+1}+e^{x+1}& 4x{e}^x-e^x& x{e}^x+e^x\\ x{e}^{x+1}+2e^{x+1}& 4x{e}^x+3e^x& x{e}^x+2e^x\end{array}\right|$

Then we have

$$\begin{gathered} W\left(f_1\left(x\right),f_2\left(x\right),f_3\left(x\right)\right)=\left(x{e}^{x+1}\right)\times \left|\begin{array}{cc}4x{e}^x-e^x& x{e}^x+e^x\\ 4x{e}^x+3e^x& x{e}^x+2e^x\end{array}\right| \\ =-\left(4x-5\right)e^x\times \left|\begin{array}{cc}x{e}^{x+1}+e^{x+1}& x{e}^x+e^x\\ x{e}^{x+1}+2e^{x+1}& x{e}^x+2e^x\end{array}\right|+\left(x{e}^x\right)\times \left|\begin{array}{cc}x{e}^{x+1}+e^{x+1}& 4x{e}^x-e^x\\ x{e}^{x+1}+2e^{x+1}& 4x{e}^x+3e^x\end{array}\right| \end{gathered}$$

Simplifying the equations

Then we have

$$\begin{gathered} W\left(f_1\left(x\right),f_2\left(x\right),f_3\left(x\right)\right)=x{e}^{x+1}\times \left(4x^2{e}^{2x}+8x{e}^{2x}-x{e}^{2x}-2e^{2x}-4x^2{e}^{2x}-4x{e}^{2x}-3x{e}^{2x}-3e^x+\left(5-4x\right)x^x\right) \\ \times \left(x^2{e}^{2x+1}+2x{e}^{2x+1}+x{e}^{2x+1}+2e^{2x+1}-x^2{e}^{2x+1}-x{e}^{2x+1}-2x{e}^{2x+1}-2e^{2x+1}\right)+x{e}^x\times \left(4x^2{e}^{2x+1}+3x{e}^{2x+1}+4x{e}^{2x+1}+3e^{2x+1}-4x^2{e}^{2x+1}+x{e}^{2x+1}-8x{e}^{2x+1}+2e^{2x+1}\right) \end{gathered}$$

Then

$$\begin{gathered} W\left(f_1\left(x\right),f_2\left(x\right),f_3\left(x\right)\right)=x{e}^{x+1}\times \left(-5e^{2x}\right)+\left(5-4x\right)e^x\times \left(0\right)+x{e}^x\times \left(5e^{2x+1}\right) \\ =-5x{e}^{3x+1}+5x{e}^{3x+1} \\ =0 \end{gathered}$$

Since we have $W\left(f_1\left(x\right),f_2\left(x\right),f_3\left(x\right)\right)=0$, then this set of functions is a linearly dependent.