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

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- Aug 22nd 2011, 04:11 AMjonnyldst 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 AMProve ItRe: dst differential equation
$\displaystyle \displaystyle \begin{align*} \frac{d^2x}{dt^2} &= -\frac{dx}{dt} + e^{-t} + 3 \\ \frac{d^2x}{dt^2} + \frac{dx}{dt} &= e^{-t} + 3 \end{align*}$

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 AMmr fantasticRe: dst differential equation