# Simple system of two nonhomogeneous ordinary differential equations solved by elimination. (3.1-15)

My differential equations textbook states to use the "elimination method" to crack this for $x$ and $y$. The final answer uses $t$ as the independent variable which both $x$ and $y$ are dependent on. I was able to solve this for $x(t)$ but it is $y(t)$ where I am having difficulty duplicating the answer in text.

The system consists of the following two linear ordinary differential equations written in linear differential operator forms:

\begin{align*} &(1) \: D(x + y) = x + t &\\ &(2) \: D^2y = Dx & \end{align*}

The textbook states that the general solution for both $x$ and $y$ are:

\begin{align*} & x(t) = \frac{1}{2}t^2 + c_1t + c_2 &\\ & y(t) = \frac{1}{6}t^3 + \frac{1}{2}c_1t^2 + (c_2 - c_1)t + c_3 &\\ \end{align*}

Again I had no problem deriving $x(t)$. It is my solution for $y(t)$ which almost but not fully agrees with the answer in the text. The following steps shows my derivation for $y(t)$.

I start by differentiating the known solution for $x$:

\begin{align*} & x = \frac{1}{2}t^2 + c_1t + c_2 &\\ & x' = t + c_1 &\\ \end{align*}

We now integrate twice to find $y$ with substitutions (shown in parenthesis) along the way for $x'$ and $x$ while combining arbitrary constants as needed yielding:

\begin{align*} & \: y'' = x' &\\ & \: y'' = (t + c_1) &\\ & \: y' = x + c &\\ & \: y' = (\frac{1}{2}t^2 + c_1t + c_2)+ c &\\ & \: y' = \frac{1}{2}t^2 + c_1t + c_2 \\ & \: y = \frac{1}{6}t^3 + \frac{1}{2}c_1t^2 + c_2t + c_3 & \end{align*}

Again my textbook states that the general solution for $y$ should look like this:

\begin{align*} & y(t) = \frac{1}{6}t^3 + \frac{1}{2}c_1t^2 + (c_2 - c_1)t + c_3 &\\ \end{align*}

But I get this:

\begin{align*} & \: y(t) = \frac{1}{6}t^3 + \frac{1}{2}c_1t^2 + c_2t + c_3 & \end{align*}

Where did the $-c_1t$ come from?

I'm not sure what that elimination method entails, probably it is expected that you first differentiate the first equation to then eliminate $y$ from it.
However, the second equation can directly be integrated once to give $$Dy=x+c_1$$ which inserted into the first equation gives $$D(x+y)=Dx+x+c_1=x+t$$ which reduces to $$Dx = t-c_1 \implies x=\frac12t^2-c_1t+c_2$$ and inserting backwards $$Dy=x+c_1=\frac12t^2-c_1t+c_1+c_2\implies y=\frac16t^3-\frac12c_1t^2+(c_1+c_2)t+c_3$$ Changing the sign of the constant $c_1$ gives the text book answer.