Find y0 so that the integral curve for dy/dx=x+y y(-4)=y0 is a straight line. You must justify your answer, which will require you to apply algebraic reasoning to the problem.
I know the answer is 3. I also know this differential equation looks simple, but I can't get the starting equation separated. We are studying separable equations so I must do it using that method. If someone could help me get it separated I think I can handle it. Thank you.
I'm feeling an urge to do a change of variables: y = t*x
Firstly, the DE y' = x + y can be solved like such:
Multiply by e-x: e-xy' - e-xy = xe-x
Integrate: e-xy = -e-x(x+1) + c1
Solve; divide by e-x: y(x) = c1ex - x - 1
Secondly, solve for c1.
IVP: y(-4) = y0 = c1e-4 + 3
Solve: c1 = (y0 - 3)e4
Input: y(x) = (y0 - 3)ex+4 - x - 1
When y0 - 3, the equation is y(x) = -x - 1.
This can be reasoned as the only solution because while y = -x - 1 by definition a strait line, y=cex+4 is not a strait line except at c=0. There are, of course, other mathematical ways of proving this.