y\'1 = 9y1+13.5y2 ------1 y\'2 = 1.5y1+9y2 -------2 The answer is y1 = 3c1e^13.5

ID: 2982903 • Letter: Y

Question

y'1 = 9y1+13.5y2 ------1

y'2 = 1.5y1+9y2 -------2

The answer is

y1 = 3c1e^13.5t +3c2e^4.5t

y2 = c1e^13.5t - c2e^4.5t

I get the answer with the matrix form, but I want to solve it this way.....

y2 = y'[1/13.5] -[9/13.5]y1

y'2 = [1/13.5]y'' - [9/13.5]y1' ----we plug in this equation to equation 2

[1/13.5]y'' - [9/13.5]y1' = 1.5y1+ 9[(1/13.5)y' - (9/13.5)y1]

after doing the rest of the arithmetic I get

the following

y'' -18y' +60.75 = 0

y1 = Ce^4.5t +De^13.5t ------------3

and to get y2 we use take derivative of equation 3 and plug it into equation 1

y'1 = 4.5ce^4.5t +13.5De^13.5t

4.5c^4.5t +13.5De^13.5t = 9[Ce^4.5t +De^13.5t ] +13.5y2

13.5y2 = -4.5Ce^4.5t+4.5De^13.5t ; then we divide everything by 13.5 to get y2 equation

y2 = -0.33Ce^4.5t +0.333De^13.5t

I just can't see how we get the answer I gave above of

y1 = 3c1e^13.5t +3c2e^4.5t

y2 = c1e^13.5t - c2e^4.5t

-0.33Ce^4.5t +0.333De^13.5t

is the same asc1e^13.5t - c2e^4.5t

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.