# Thread: Particular Solution of the differential equation

1. ## Particular Solution of the differential equation

Determine :- Particular Solution of the differential equation:

(12)d^2y/dt2 - 3y = 0

Boundaries t=0, y=3, dy/dt=0.5

Thanks

2. ## Re: Particular Solution of the differential equation

what have you tried?

where are you stuck?

3. ## Re: Particular Solution of the differential equation

Are you taking a course in differential equations? This is a very basic, simple example. Its "characteristic equation" is $\displaystyle 12r^2- 3= 0$ which has roots $\displaystyle r= \pm\frac{1}{2}$ so the "general solution" to the differential equation is $\displaystyle y(t)= Ae^{t/2}+ Be^{-t/2}$ (or equivalently, $\displaystyle y(t)= C cosh(t/2)+ D sinh(t/2)$).

From the formula $\displaystyle y(t)= Ae^{t/2}+ Be^{-t/2}$, $\displaystyle y'(t)= (A/2)e^{t/2}- (B/2)e^{-t/2}$ so y(0)= A+ B= 3 and y'(0)= A/2- B/2= 1/2.

From the formula $\displaystyle y(t)= C cosh(t/2)+ D sinh(t/2)$, $\displaystyle y'(t)= (C/2)sinh(t/2)+ (D/2)cosh(t/2)$ so y(0)= C= 3 and y'(0)= D/2= 1/2.

4. ## Re: Particular Solution of the differential equation

Originally Posted by HallsofIvy
Are you taking a course in differential equations? This is a very basic, simple example. Its "characteristic equation" is $\displaystyle 12r^2- 3= 0$ which has roots $\displaystyle r= \pm\frac{1}{2}$ so the "general solution" to the differential equation is $\displaystyle y(t)= Ae^{t/2}+ Be^{-t/2}$ (or equivalently, $\displaystyle y(t)= C cosh(t/2)+ D sinh(t/2)$).

From the formula $\displaystyle y(t)= Ae^{t/2}+ Be^{-t/2}$, $\displaystyle y'(t)= (A/2)e^{t/2}- (B/2)e^{-t/2}$ so y(0)= A+ B= 3 and y'(0)= A/2- B/2= 1/2.

From the formula $\displaystyle y(t)= C cosh(t/2)+ D sinh(t/2)$, $\displaystyle y'(t)= (C/2)sinh(t/2)+ (D/2)cosh(t/2)$ so y(0)= C= 3 and y'(0)= D/2= 1/2.
changed from "math /math" to "tex /tex" delimiters

5. ## Re: Particular Solution of the differential equation

Sorry, I was unable to attach the image of my workings.. For some reason I thought the root was a complex number (j0.5). Which resulted in using the solution y= e^ αx (Acosβx+BsinβX). I now see it not a complex number to the general solution is what Romsek provided.

Thanks!

6. ## Re: Particular Solution of the differential equation

Originally Posted by mrdannydandan
Sorry, I was unable to attach the image of my workings.. For some reason I thought the root was a complex number (j0.5). Which resulted in using the solution y= e^ αx (Acosβx+BsinβX). I now see it not a complex number to the general solution is what Romsek provided.

Thanks!
what HallsOfIvy provided. I just got his LaTex working.

7. ## Re: Particular Solution of the differential equation

oh yes... Thank you HallsOfIvy