# dst differential equation

• Aug 22nd 2011, 04:11 AM
jonnyl
dst differential equation
I've got an answer for this question but i'd like to check it against someone else and see the working used
• Aug 22nd 2011, 04:41 AM
Prove It
Re: dst differential equation
Quote:

Originally Posted by jonnyl
I've got an answer for this question but i'd like to check it against someone else and see the working used

\displaystyle \displaystyle \begin{align*} \frac{d^2x}{dt^2} &= -\frac{dx}{dt} + e^{-t} + 3 \\ \frac{d^2x}{dt^2} + \frac{dx}{dt} &= e^{-t} + 3 \end{align*}

You need to solve the homogeneous equation $\displaystyle \displaystyle \frac{d^2x}{dt^2} + \frac{dx}{dt} = 0$.

The characteristic equation is

\displaystyle \displaystyle \begin{align*} m^2 + m &= 0 \\ m(m + 1) &= 0 \\ m = 0 \textrm{ or }m &= -1 \end{align*}

which means the solution to the homogeneous equation is

$\displaystyle \displaystyle x_c = Ae^{0t} + Be^{-t} = A + Be^{-t}$.

Now, using the method of undetermined coefficients to solve the nonhomogeneous equation, guess $\displaystyle \displaystyle x_p = C_1t^2 + C_2t + C_3 + C_4t\,e^{-t}$, then $\displaystyle \displaystyle \frac{dx}{dt} = 2C_1t + C_2 + C_4e^{-t} - C_4t\,e^{-t}$ and $\displaystyle \displaystyle \frac{d^2x}{dt^2} = 2C_1 - C_4e^{-t} - C_4e^{-t} + C_4t\,e^{-t}$.

Substitute each of these into the DE $\displaystyle \displaystyle \frac{d^2x}{dt^2} + \frac{dx}{dt} = e^{-t} + 3$ and equate like coefficients to solve for $\displaystyle \displaystyle C_1, C_2, C_3, C_4$.
• Aug 22nd 2011, 04:50 AM
mr fantastic
Re: dst differential equation
Quote:

Originally Posted by jonnyl
I've got an answer for this question but i'd like to check it against someone else and see the working used

If you have an answer, you can check it simply by substituting it into the DE and seeing if it works.