Separable Equation (t^2 -12t+27)dy/dt=y

[13teresap]

Hello,

I am new to this website, so please bare with me.

I am having trouble completing this equation.

. . . . .\(\displaystyle \mbox{Solve }\, (t^2\, -\, 12t\, +\, 27)\, \dfrac{dy}{dt}\, =\, y\, \mbox{ when }\, y(6)\, =\, 1\)

I have tried multiple ways but each time I find a function for y(t), I end up needing take the root or natural log of a number. Which I don't believe is possible for the Differential Equations class I am currently in...

So this is what I have so far ( I hope this is making sense...).

. . . . .\(\displaystyle \dfrac{y'}{y}\, =\, \dfrac{1}{t^2\, -\, 12t\, +\, 27}\)

**I tried to factor the polynomial for so I could integrate it easier? Might have made it worse**

. . . . .\(\displaystyle \dfrac{y'}{y}\, =\, \left(\dfrac{1}{6}\right)\, \left[\, \left(\dfrac{1}{t\, -\, 9}\right)\, -\, \left(\dfrac{1}{t\, -\, 3}\right)\, \right]\)

**Then I integrated both sides**

. . . . .\(\displaystyle \ln(y)\, =\, \left[\, \bigg(\dfrac{1}{6}\right)\, \left(\ln(t\, -\, 9)\, -\, \ln(t\, -\, 3)\bigg)\, \right]\, +\, C\)

**Then I tried to cancel out the natural logs with e**

. . . . .\(\displaystyle y\, =\, \dfrac{e^C\, (t\, -\, 9)^{1/6}}{(t\, -\, 3)^{1/6}}\)

**This where I run into my problem**

If I input my t value of 6, I get a negative number for my 6th root in the numerator of my problem which is preventing me from finding a value for C...

I really hope someone can understand this and help me...

 

[stapel]

First off, please let me say a big "thank you!" for showing your work and reasoning so clearly! :cool:

I am having trouble completing this equation.

. . . . .\(\displaystyle \mbox{Solve }\, (t^2\, -\, 12t\, +\, 27)\, \dfrac{dy}{dt}\, =\, y\, \mbox{ when }\, y(6)\, =\, 1\)

I have tried multiple ways but each time I find a function for y(t), I end up needing take the root or natural log of a number. Which I don't believe is possible for the Differential Equations class I am currently in...

So this is what I have so far ( I hope this is making sense...).

. . . . .\(\displaystyle \dfrac{y'}{y}\, =\, \dfrac{1}{t^2\, -\, 12t\, +\, 27}\)

**I tried to factor the polynomial for so I could integrate it easier? Might have made it worse**

. . . . .\(\displaystyle \dfrac{y'}{y}\, =\, \left(\dfrac{1}{6}\right)\, \left[\, \left(\dfrac{1}{t\, -\, 9}\right)\, -\, \left(\dfrac{1}{t\, -\, 3}\right)\, \right]\)

**Then I integrated both sides**

. . . . .\(\displaystyle \ln(y)\, =\, \left[\, \bigg(\dfrac{1}{6}\right)\, \left(\ln(t\, -\, 9)\, -\, \ln(t\, -\, 3)\bigg)\, \right]\, +\, C\)

Technically, the logs should be of absolute values. (here) (And, yes, you're right: we hardly ever care about this. However, yours is one of the particular cases in which this can matter.)

**Then I tried to cancel out the natural logs with e**

. . . . .\(\displaystyle y\, =\, \dfrac{e^C\, (t\, -\, 9)^{1/6}}{(t\, -\, 3)^{1/6}}\)

Wolfram Alpha agrees with you. (here) (They don't state their answer in exactly the same form, but the results are equivalent.)

**This where I run into my problem**

If I input my t value of 6, I get a negative number for my 6th root in the numerator of my problem which is preventing me from finding a value for C...

So let's use the absolute values, instead:

. . . . .\(\displaystyle \ln(y)\, =\, \left[\, \bigg(\dfrac{1}{6}\right)\, \left(\ln\left|\strut t\, -\, 9 \right| \, -\, \ln\left|\strut t\, -\, 3 \right| \bigg)\, \right]\, +\, C\)

This leads to:

. . . . .\(\displaystyle y\, =\, e^C\, \sqrt[6]{\strut \left| \dfrac{t\, -\, 9}{t\, -\, 3} \right|\, }\)

. . . . .\(\displaystyle 1\, =\, e^C\, \sqrt[6]{\strut \left| \dfrac{-3}{3} \right|\, }\, =\, e^C\, \sqrt[6]{\strut 1\,}\, =\, e^C\)

. . . . .\(\displaystyle \ln(1)\, =\, C\)

...and so forth.

 

[13teresap]

Thank you!

Stapel,

Thank you so much!

I did exactly what you had mentioned, I ignored the absolute value rule when working with natural logs.

I will remember this next time!