Question

Differential Equations

Solve the given initial value problem. y'" - 2y" - 36y' + 72y = 0 y(O)= -13, y'(O)= - 34y''(0) = - 308 y(x) = 0

Answer 1

The given differential equation is $y'''-2y''-36y'+72=0$

with initial conditions

y (0) = -13, y0) = -34, y"0) = -308

This is a homogenous differential equation of 3rd order with constant coefficients. The auxiliary equation is

m– 2m? – 36m + 72 = 0 = (m - 2)(m? – 36) = 0 2,6, -6 m

Hence the general solution is

61 y(2) = Cie -6r Cze + C3e

where $c_1,c_2,c_3$ are constants which can be determined using initial conditions.

If the general solution satisfies the initial condition, then we obtain the following equations

$c_1+c_2+c_3=-13$

$2c_1-6c_2+6c_3=-34$

$4c_1+36c_2+36c_3=-308$

We shall solve this linear system of equations by matrix row-echelon method with augmented matrix

$\begin{pmatrix}1&1&1&-13\\ \:2&-6&6&-34\\ \:4&36&36&-308\end{pmatrix}$

$R_2\rightarrow R_2-2R_1$, $R_3\rightarrow R_3-4R_1$

$\begin{pmatrix}1&1&1&-13\\ \:0&-8&4&-8\\ \:0&32&32&-256\end{pmatrix}$

$R_2\rightarrow \frac{R_2}{4},\;R_3\rightarrow \frac{R_3}{32}$

$\begin{pmatrix}1&1&1&-13\\ \:0&-2&1&-2\\ \:0&1&1&-8\end{pmatrix}$

$R_3\rightarrow 2R_3+R_2$

$\begin{pmatrix}1&1&1&-13\\ \:\:0&-2&1&-2\\ \:\:0&0&3&-18\end{pmatrix}$

This gives

$c_1=-5,\;c_2=-2,\;c_3=-6$

hence general solution is

$y(x)=-5e^{2x}-2e^{-6x}-6e^{6x}$