Answer:
[tex]y=\frac{17}{13} e^{-x}+\frac{48}{13} e^{\frac{1}{2}x }\cos(x)+\frac{6}{13} e^{\frac{1}{2}x }\sin(x)[/tex]
Step-by-step explanation:
Given the third-order differential equation with initial conditions.
[tex]4y'''+y'+5y=0; \ y(0)=5, \ y'(0)=1, \ y''(0)=-1[/tex]
(1) - Find the characteristic equation
[tex]4y'''+y'+5y=0\\\\\Longrightarrow 4m^3+0m^2+m+5=0\\\\\Longrightarrow \boxed{4m^3+m+5=0}[/tex]
(2) - Solve the characteristic equation for "m." First using the rational root theorem
[tex]4m^3+m+5=0\\\\\rightarrow \frac{p}{q} = \pm \frac{1,5}{1,2,4} \\\\\text{\underline{Trying m=-1 first: }}\\\\ \begin{array}{c|rrrr}\vphantom{\dfrac12}-1 & 4 & 0 & 1 & 5\\\cline{1-1} & \downarrow & -4 & 4 & -5\\\cline{2-5} & 4 & -4 & 5 & 0\end{array}\\\\\text{The remainder is 0} \ \therefore \boxed{m=-1} \ \text{is a root.}\\\\\text{Now we are left with}\rightarrow \boxed{4m^2-4m+5=0} \ \text{Use the quadratic equation}[/tex]
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{The Quadratic Equation:}}\\ax^2+bx+c=0\\\\x=\frac{-b \pm\sqrt{b^2-4ac} }{2a} \end{array}\right}[/tex]
[tex]4m^2-4m+5=0\\\\\Longrightarrow m=\frac{4 \pm\sqrt{(-4)^2-4(4)(5)} }{2(4)} \\\\\Longrightarrow m=\frac{4 \pm\sqrt{-64} }{8}\\\\\Longrightarrow m=\frac{4}{8} \pm \frac{8i}{8} \\\\\Longrightarrow \boxed{m=\frac{1}{2} \pm i}[/tex]
Thus, we have found three roots.
[tex]m=-1, \frac{1}{2} \pm i[/tex]
(3) - Form the solution.
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Solutions to Higher-order DE's:}}\\\\\text{Real,distinct roots} \rightarrow y=c_1e^{m_1t}+c_2e^{m_2t}+...+c_ne^{m_nt}\\\\ \text{Duplicate roots} \rightarrow y=c_1e^{mt}+c_2te^{mt}+...+c_nt^ne^{mt}\\\\ \text{Complex roots} \rightarrow y=c_1e^{\alpha t}\cos(\beta t)+c_2e^{\alpha t}\sin(\beta t)+... \ ;m=\alpha \pm \beta i\end{array}\right}[/tex]
Notice that we have one real, distinct root and complex roots. Thus, we can form the solution in the following manner.
[tex]\therefore \text{the general solution}\rightarrow\boxed{\boxed{y=c_1e^{-x}+c_2e^{\frac{1}{2}x }\cos(x)+c_3e^{\frac{1}{2}x }\sin(x)}}[/tex]
(4) - Use the given initial conditions to find the arbitrary constants "c_1," "c_2," and "c_3"
[tex]\text{Recall...} \ y(0)=5, \ y'(0)=1, \ y''(0)=-1[/tex]
Take two derivatives of the general solution.
[tex]y=c_1e^{-x}+c_2e^{\frac{1}{2}x }\cos(x)+c_3e^{\frac{1}{2}x }\sin(x)\\\\\\ \Rightarrow y'=-c_1e^{-x}+c_2\frac{1}{2}e^{\frac{1}{2}x}\cos(x)-c_2e^{\frac{1}{2}x}\sin(x)+c_3e^{\frac{1}{2}x }\cos(x)+c_3\frac{1}{2}e^{\frac{1}{2}x }\sin(x) \\\\\Longrightarrow \boxed{y'=-c_1e^{-x}+(c_3+\frac{1}{2} c_2)e^{\frac{1}{2}x}\cos(x)+(-c_2+\frac{1}{2} c_3)\frac{3}{2}e^{\frac{1}{2}x }\sin(x)}\\\\\\[/tex]
[tex]\Rightarrow y''=c_1e^{-x}+\frac{1}{2} (c_3+\frac{1}{2} c_2)e^{\frac{1}{2}x }\cos(x)+(c_3+\frac{1}{2} c_2) e^{\frac{1}{2}x}\sin{x}+(-c_2+\frac{1}{2} c_3)e^{\frac{1}{2}x } \cos(x)+\frac{1}{2} (-c_2+\frac{1}{2} c_3)e^{\frac{1}{2}x }\sin(x)\\\\\Longrightarrow \boxed{y''=c_1e^{-x}+(-\frac{3}{4} c_2+c_3)e^{\frac{1}{2}x }\cos(x)+(- c_2-\frac{3}{4}c_3) c_3e^{\frac{1}{2}x }\sin(x)}[/tex]
Plug in the initial conditions and form a system of equations.
[tex]\left\{\begin{array}{ccc}5=c_1+c_2\\1=-c_1+\frac{1}{2}c_2+c_3 \\-1=c_1-\frac{3}{4}c_2+c_3 \end{array}\right[/tex]
Creating a matrix and using a calculator to row-reduce,
[tex]\Longrightarrow\left[\begin{array}{ccc}1&1&0\\-1&\frac{1}{2}&1\\1&-\frac{3}{4}&1\end{array}\right] =\left[\begin{array}{ccc}5\\1\\-1\end{array}\right] \\\\\Longrightarrow \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] =\left[\begin{array}{ccc}\frac{17}{13} \\\frac{48}{13} \\\frac{6}{13} \end{array}\right] \\\\\therefore \boxed{c_1=\frac{17}{13} , c_2=\frac{48}{13} , \ and \ c_3=\frac{6}{13} }[/tex]
(5) - Thus, the given differential equation is solved with the given initial conditions
[tex]\boxed{\boxed{y=\frac{17}{13} e^{-x}+\frac{48}{13} e^{\frac{1}{2}x }\cos(x)+\frac{6}{13} e^{\frac{1}{2}x }\sin(x)}}[/tex]
