Method of undetermined coefficients

**alexmahone** · November 29th 2011, 10:26 AM

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

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

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

**Prove It** · November 29th 2011, 03:31 PM

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 $t$ .

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

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

1) From the first equation

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

and from the second

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

Therefore

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

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

**alexmahone** · November 29th 2011, 03:33 PM

Originally Posted by Prove It

Am I correct in assuming that you're using $\displaystyle \begin{align*} x' \end{align*}$ to represent $\displaystyle \begin{align*} \frac{dx}{dt} \end{align*}$ and $\displaystyle \begin{align*} y' \end{align*}$ to represent $\displaystyle \begin{align*} \frac{dy}{dt} \end{align*}$ ...

Yes. (I said that in my second sentence.

)

**Prove It** · November 29th 2011, 03:45 PM

Originally Posted by alexmahone

Yes. (I said that in my second sentence.

)

I edited my post.

**alexmahone** · November 29th 2011, 03:53 PM

Originally Posted by Prove It

1) From the first equation

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

and from the second

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

Therefore

$\displaystyle \begin{align*} \frac{d^2x}{dt^2} - \frac{d^2y}{dt^2} &= \frac{dx}{dt} + \frac{dy}{dt} + 2 - \left( \frac{dx}{dt} + \frac{dy}{dt} - 2 \right) \\ \frac{d^2x}{dt^2} + \frac{d^2y}{dt^2} &= 4\end{align*}$

Shouldn't that last equation be $\frac{d^2x}{dt^2}-\frac{d^2y}{dt^2}=4$ ?

**Prove It** · November 29th 2011, 03:59 PM

Originally Posted by alexmahone

Shouldn't that last equation be $\frac{d^2x}{dt^2}-\frac{d^2y}{dt^2}=4$ ?

Yes it should be, sorry, so there'll be sign errors through the rest of the working, but the process still works

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