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.

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

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

Recall that the roots of a polynomial of the form are given by the formula

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

So, in this case, the general solution is

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

(or equivalently

By replacing the second equation in the first one, we get which implies that .

So

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

(or equivalently

By solving this system, the solution is

Then

c)

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

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

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.