# Thread: Method of undetermined coefficients

1. ## Method of undetermined coefficients

Apply the method of undetermined coefficients to find a particular solution of each of the systems. Primes denote derivatives with respect to $\displaystyle t$.

1) $\displaystyle x'=x+y+2t,\ y'=x+y-2t$

2) $\displaystyle x'=2x+y+1,\ y'=3x+2y+e^{4t}$

2. ## Re: Method of undetermined coefficients

Originally Posted by alexmahone
Apply the method of undetermined coefficients to find a particular solution of each of the systems. Primes denote derivatives with respect to $\displaystyle t$.

1) $\displaystyle x'=x+y+2t,\ y'=x+y-2t$

2) $\displaystyle x'=2x+y+1,\ y'=3x+2y+e^{4t}$
1) From the first equation

\displaystyle \displaystyle \begin{align*} \frac{dx}{dt} &= x + y + 2t \\ \frac{d^2x}{dt^2} &= \frac{dx}{dt} + \frac{dy}{dt} + 2 \end{align*}

and from the second

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

Therefore

\displaystyle \displaystyle \begin{align*} \frac{d^2x}{dt^2} - \frac{d^2y}{dt^2} &= \frac{dx}{dt} + \frac{dy}{dt} + 2 - \left( \frac{dx}{dt} + \frac{dy}{dt} - 2 \right) \\ \frac{d^2x}{dt^2} + \frac{d^2y}{dt^2} &= 4 \\ \frac{d^2y}{dt^2} &= 4 - \frac{d^2x}{dt^2} \\ \frac{dy}{dt} &= 4t - \frac{dx}{dt} + C_1 \\ y &= 2t^2 - x + C_1t + C_2 \\ x + y &= 2t^2 + C_1t + C_2 \\ \frac{dx}{dt} - 2t &= 2t^2 + C_1t + C_2 \\ \frac{dx}{dt} &= 2t^2 + (C_1 + 2)t + C_2 \\ x &= \frac{2}{3}t^3 + \left(\frac{C_1 + 2}{2}\right)t^2 + C_2t + C_3 \end{align*}

and from a similar process you can find \displaystyle \displaystyle \begin{align*} y \end{align*}

3. ## Re: Method of undetermined coefficients

Originally Posted by Prove It
Am I correct in assuming that you're using \displaystyle \displaystyle \begin{align*} x' \end{align*} to represent \displaystyle \displaystyle \begin{align*} \frac{dx}{dt} \end{align*} and \displaystyle \displaystyle \begin{align*} y' \end{align*} to represent \displaystyle \displaystyle \begin{align*} \frac{dy}{dt} \end{align*}...
Yes. (I said that in my second sentence. )

4. ## Re: Method of undetermined coefficients

Originally Posted by alexmahone
Yes. (I said that in my second sentence. )
I edited my post.

5. ## Re: Method of undetermined coefficients

Originally Posted by Prove It
1) From the first equation

\displaystyle \displaystyle \begin{align*} \frac{dx}{dt} &= x + y + 2t \\ \frac{d^2x}{dt^2} &= \frac{dx}{dt} + \frac{dy}{dt} + 2 \end{align*}

and from the second

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

Therefore

\displaystyle \displaystyle \begin{align*} \frac{d^2x}{dt^2} - \frac{d^2y}{dt^2} &= \frac{dx}{dt} + \frac{dy}{dt} + 2 - \left( \frac{dx}{dt} + \frac{dy}{dt} - 2 \right) \\ \frac{d^2x}{dt^2} + \frac{d^2y}{dt^2} &= 4\end{align*}
Shouldn't that last equation be $\displaystyle \frac{d^2x}{dt^2}-\frac{d^2y}{dt^2}=4$?

6. ## Re: Method of undetermined coefficients

Originally Posted by alexmahone
Shouldn't that last equation be $\displaystyle \frac{d^2x}{dt^2}-\frac{d^2y}{dt^2}=4$?
Yes it should be, sorry, so there'll be sign errors through the rest of the working, but the process still works