We are learning how to solve systems of linear ODE's by elimination. I am not sure how to solve this:

x'=-y+t
y'=x-t

I took the derivative of the second eqn to get
y''-x'=-t'
and added the two to get
y''+y=-t'+t
but I don't think thats the right way to proceed. Suggestions?
tia

x'=-y+t

y'=x-t

Add these two equations, you get:

x' + y' = x - y

y' + (x')y = x

That gives you a linear differential equation of the form:

y' + Py = Q

Considering that P is the derivative of x, it'll make things easy while solving for the integrating factor.

The equation should be:

x' + y' = x - y
=> y' + (1)y = (x - x')

To solve systems of linear ODEs by elimination, we need to eliminate one variable at a time from the system. Here's how we can proceed:

1. Start with the given system of ODEs:
x' = -y + t
y' = x - t

2. To eliminate the variable x, differentiate the second equation with respect to t:
y'' = x' - 1

3. Substitute this expression for x' into the first equation:
x' = -y + t
y'' = (-y + t) - 1

4. Simplify the equation:
y'' = -y + t - 1

5. Now, we have a second-order linear differential equation for y with no x term. This equation is homogeneous since the right-hand side is zero. We can solve this equation by assuming a solution of the form y = e^(rt), where r is a constant.

6. Differentiate y twice with respect to t:
y' = re^(rt)
y'' = r^2e^(rt)

7. Substitute these expressions into the differential equation:
r^2e^(rt) = -e^(rt) + t - 1

8. Divide the entire equation by e^(rt) to remove the exponential term:
r^2 = -1 + te^(-rt) - e^(-rt)

9. Multiply through by e^(rt) to simplify the equation:
r^2e^(rt) = -e^(2rt) + te^(-rt) - 1

10. Now we have a nonlinear equation for r. This equation can be solved numerically, graphically, or using approximation techniques (such as Taylor series or power series expansions).

11. Once you have found the value(s) of r, substitute them back into the equation y = e^(rt) to obtain the solutions for y.

12. Finally, substitute the solutions for y back into the original equations to solve for x.

It's worth noting that this method may not always yield a closed-form solution, and numerical methods may be required to approximate the solutions.