Differential Equation..help please..
6y"+y'-y=0, y sub 1 = e^(-x/2). I am getting [(-1+/-(sqrt 23)i]/12 using quadratic eqtn. Where does the given ysub1 become relevant?

Question

nikvist · Answer

$$6y''+y'-y=0$$$$6\lambda^2+\lambda-1=0$$$$\lambda_{1,2}=\frac{-1\pm\sqrt{1+24}}{12}=\frac{-1\pm 5}{12}$$$$\lambda_1=-1/2\quad,\quad \lambda_2=1/3$$$$y=C_1\exp{(-x/2)}+C_2\exp{(x/3)}$$

anonymous · Answer

under the radical for quadratic, it's b^2 minus 4AC, not plus. Am I wrong?

anonymous · Answer

oh nm it started as a negative!

anonymous · Answer

so this given y sub 1 is just one of the C sub 1's? Just confused on notation. Thanks a ton.

kira_yamato · Answer

I'll use the complex plane instead... Assuming I'm taking derivatives wrt t.

anonymous · Answer

The instruction said to construct a second solution for the given DE. They gave y sub 1 = e^(-x/2). Is there more work than just finding 1/3 and -1/2?

kira_yamato · Answer

Sorry.....

anonymous · Answer

Anybody? Using the y sub 1 to construct a 2nd solution...?