Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is a linear combination of the two independent solutions of this differential equation that you found first. You are not being asked for just one of these. You will need to determine the values of the two constant parameters c1 and c2. Similarly for finding y2 below. Find the function y2 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y2(0)=0,y′2(0)=1. y2= Find the Wronskian W(t)=W(y1,y2). W(t)= Remark: You can find W by direct computation and use Abel's theorem as a check. You should find that W is not zero and so y1 and y2 form a fundamental set of solutions of 121y′′+110y′−24y=0.

The original equation is [tex]121y''+110y'-24y=0[/tex]. We propose that the solution of this equations is of the form [tex] y = Ae^{rt}[/tex]. Then, by replacing the derivatives we get the following

[tex]121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)[/tex]

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

[tex]121r^2+110r-24=0[/tex]

Recall that the roots of a polynomial of the form [tex]ax^2+bx+c[/tex] are given by the formula

[tex] x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}[/tex]

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

[tex]r_1 = -\frac{12}{11}[/tex]

[tex]r_2 = \frac{2}{11}[/tex]

So, in this case, the general solution is [tex]y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}[/tex]

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

[tex]c_1 + c_2 = 1[/tex]

[tex]c_1\frac{-12}{11} + c_2\frac{2}{11} = 0[/tex](or equivalently [tex]c_2 = 6c_1[/tex]

By replacing the second equation in the first one, we get [tex]7c_1 = 1 [/tex] which implies that [tex] c_1 = \frac{1}{7}, c_2 = \frac{6}{7}[/tex].

So [tex]y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}[/tex]

b) By using y(0) =0 and y'(0)=1 we get the equations

[tex] c_1+c_2 =0[/tex]

[tex]c_1\frac{-12}{11} + c_2\frac{2}{11} = 1[/tex](or equivalently [tex]-12c_1+2c_2 = 11[/tex]

By solving this system, the solution is [tex]c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}[/tex]

Then [tex]y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}[/tex]

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

[tex]\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[/tex]

By plugging the values of [tex]y_1[/tex] 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

[tex]e^{\int -p(x) dx}[/tex]

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

[tex]e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}[/tex]

Note that this function is always positive, and thus, never zero. So [tex]y_1, y_2[/tex] is a fundamental set of solutions.