Differential Equations

Question

Solve the following initial value problem:

(t^2-16t+60)dy/dt = y

 

1) with y(8)=1y(8)=1. (Find yy as a function of t).

2) On what interval is the solution valid?

3) Find the limit of the solution as tt approaches the left end of the interval.

4) Similar to 3), but for the right end.

 

Answer 1

A quick partial solution looks like this:

 

dy/y = dt/(t^2 - 16t + 60)  <---- separation of variables

dy/y = dt/{(t - 10)(t-6)}  <---- factors

ln|y| = 1/4 * ln | 4/(t-6)-1|+ c

<--- integration done by partial fraction decomposition as follows ---->

A/(t-10) + B/(t-6) = 1/((t-10)(t-6))

A(t-6) + B(t-10) = 1

At - 6A + Bt - 10B = 1

A+B = 0 ---> B = -A

-6A-10B = 1

6B - 10B = 1

B = -1/4 ---> A = 1/4

1/4/(t - 10) + -1/4/(t - 6)

1/4 ln(t-10) - 1/4 ln(t-6)

1/4 ln ( (t-10)/(t-6)) = 1/4 ( 1 - 4/t-6) )

<------------------------------------------------------------------>
Graphing the argument of the log 4/(t-6) - 1
shows positive values between 6 and 10

THe log blows up as t approaches 6