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.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}{\mathbf{e}}^{\mathbf{x}}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{,}{\mathbf{e}}^{\mathbf{x}}\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{,}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$

Short answer

The General solution is $y=c_1{e}^x\cos 2x+c_2{e}^x\sin 2x$.

Step-by-step solution

Verify the function for the given differential equation

For the first function$y_1\left(x\right)=e^x\cos 2x$, we have

$y\text{'}=e^x(\cos 2x-2\sin 2x)andy\text{'}\text{'}=e^x(-4\sin 2x-3cos2x)$

Substituting the derivatives in the DE gives,

$\begin{array}{rcl}y\text{'}\text{'}-2y\text{'}+5y& =& 0\\ e^x(4\cos 2x-3\sin 2x)-2e^x(\sin 2x+2\cos 2x)+5e^x\sin 2x& =& 0\\ e^x(4\cos 2x-3\sin 2x-2\sin 2x-4\cos 2x+5\sin 2x)& =& 0\end{array}$

Similarly for the second function$y_2\left(x\right)=e^x\sin 2x$ , we have

$y\text{'}=e^x(\sin 2x+2\cos 2x)andy\text{'}\text{'}=e^x(4\cos 2x-3\sin 2x)$

Substituting the derivatives in the DE gives,

$\begin{array}{rcl}y\text{'}\text{'}-2y\text{'}+5y& =& 0\\ e^x(4\cos 2x-3\sin 2x)-2e^x(\sin 2x+2\cos 2x)+5e^x\sin 2x& =& 0\\ e^x(4\cos 2x-3\sin 2x-2\sin 2x-4\cos 2x+5\sin 2x)& =& 0\\ 0& =& 0\end{array}$

Both the function satisfies the DE, the functions$y=e^x\cos 2xandy=e^x\sin 2x$ are the solutions of the differential equation.

Determine the set of function by using Wronskian

The coefficient functions $a_2\left(x\right)=1,a_1\left(x\right)=-2and{a}_0\left(x\right)=5$are continuous on $(-\infty ,\infty )$and that $a_2\left(x\right)\ne 0$ for every value of x in the interval. Wronskian is

$\begin{array}{rcl}W\left(e^x\cos 2x,e^x\sin 2x\right)& =& \left|\begin{array}{cc}{e}^x\cos 2x& e^x\sin 2x\\ e^x(\cos 2x-2\sin 2x)& e^x(\sin 2x+2\cos 2x)\end{array}\right|\\ & =& e^x\cos 2x·e^x(\sin 2x+2\cos 2x)-e^x\sin 2x·e^x(\cos 2x-2\sin 2x)\\ & =& e^{2x}(\left.\sin 2x\cos 2x+2{\cos }^{2}2x\right)-e^{2x}\left(\sin 2x\cos 2x-2{\sin }^{2}2x\right)\\ & =& e^{2x}\left[\sin 2x\cos 2x+2{\cos }^{2}2x\right.\left.-\sin 2x\cos 2x+2{\sin }^{2}2x\right]\\ & =& 2e^{2x}\left({\cos }^{2}2x+{\sin }^{2}2x\right)\\ & =& 2e^{2x}\end{array}$

Since $W\left(y_1\left(x\right),y_2\left(x\right)\right)=2e^{2x}\ne 0$for all x, we can conclude that$y_1\left(x\right)and{y}_2\left(x\right)$ are linearly independent on $(-\infty ,\infty )$. Hence $e^x\cos 2xand{e}^x\sin 2x$ form a fundamental set of solutions of the differential equation on the indicated interval.

Hence, the general solution of the equation on the interval is,

$\text{y}=c_1{e}^x\cos 2x+c_2{e}^x\sin 2x$

Where $c_1,c_2$are arbitrary constants.