Answer:
Step-by-step explanation:
The original equation is
. We propose that the solution of this equations is of the form
. Then, by replacing the derivatives we get the following
![121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)](https://tex.z-dn.net/?f=121r%5E2Ae%5E%7Brt%7D%2B110rAe%5E%7Brt%7D-24Ae%5E%7Brt%7D%3D0%3D%20Ae%5E%7Brt%7D%28121r%5E2%2B110r-24%29)
Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that
![121r^2+110r-24=0](https://tex.z-dn.net/?f=121r%5E2%2B110r-24%3D0)
Recall that the roots of a polynomial of the form
are given by the formula
![x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B-b%20%5Cpm%20%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions
![r_1 = -\frac{12}{11}](https://tex.z-dn.net/?f=r_1%20%3D%20-%5Cfrac%7B12%7D%7B11%7D)
![r_2 = \frac{2}{11}](https://tex.z-dn.net/?f=r_2%20%3D%20%5Cfrac%7B2%7D%7B11%7D)
So, in this case, the general solution is ![y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}](https://tex.z-dn.net/?f=y%20%3D%20c_1%20e%5E%7B%5Cfrac%7B-12t%7D%7B11%7D%7D%20%2B%20c_2%20e%5E%7B%5Cfrac%7B2t%7D%7B11%7D%7D)
a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations
![c_1 + c_2 = 1](https://tex.z-dn.net/?f=c_1%20%2B%20c_2%20%3D%201)
(or equivalently ![c_2 = 6c_1](https://tex.z-dn.net/?f=c_2%20%3D%206c_1)
By replacing the second equation in the first one, we get
which implies that
.
So ![y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}](https://tex.z-dn.net/?f=y_1%20%3D%20%5Cfrac%7B1%7D%7B7%7De%5E%7B%5Cfrac%7B-12t%7D%7B11%7D%7D%20%2B%20%5Cfrac%7B6%7D%7B7%7De%5E%7B%5Cfrac%7B2t%7D%7B11%7D%7D)
b) By using y(0) =0 and y'(0)=1 we get the equations
![c_1+c_2 =0](https://tex.z-dn.net/?f=%20c_1%2Bc_2%20%3D0)
(or equivalently ![-12c_1+2c_2 = 11](https://tex.z-dn.net/?f=-12c_1%2B2c_2%20%3D%2011)
By solving this system, the solution is ![c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}](https://tex.z-dn.net/?f=c_1%20%3D%20%5Cfrac%7B-11%7D%7B14%7D%2C%20c_2%20%3D%20%5Cfrac%7B11%7D%7B14%7D)
Then ![y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}](https://tex.z-dn.net/?f=y_2%20%3D%20%5Cfrac%7B-11%7D%7B14%7De%5E%7B%5Cfrac%7B-12t%7D%7B11%7D%7D%20%2B%20%5Cfrac%7B11%7D%7B14%7D%20e%5E%7B%5Cfrac%7B2t%7D%7B11%7D%7D)
c)
The Wronskian of the solutions is calculated as the determinant of the following matrix
![\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2](https://tex.z-dn.net/?f=%5Cleft%7C%20%5Cbegin%7Bmatrix%7Dy_1%20%26%20y_2%20%5C%5C%20y_1%27%20%26%20y_2%27%5Cend%7Bmatrix%7D%5Cright%7C%3D%20W%28t%29%20%3D%20y_1%5Ccdot%20y_2%27-y_1%27y_2)
By plugging the values of
and
We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by
![e^{\int -p(x) dx}](https://tex.z-dn.net/?f=e%5E%7B%5Cint%20-p%28x%29%20dx%7D)
In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is
Note that this function is always positive, and thus, never zero. So
is a fundamental set of solutions.