Solving Simultaneous Differential Equations

Question

I'm asking about ways to solve simultaneous differential equations. This is a question from my textbook. It asks to find $x$ and $y$ in terms of $t$: $$(1)\ \frac{dx}{dt}=3x+y$$ and $$(2)\ \frac{dy}{dt}=6x-2y$$ With the conditions that when $t=0, x=1$ and $y=15$. After differentiating $(1)$ with respect to $t$, and substituting in expressions for $y$ and $\frac{dy}{dt}$, I got a second order equation and solved it to get the general solution for $x$ as: $$(3)\ x=Ae^{4t}+Be^{-3t}$$ After this, if I differentiate this expression and use it in $(1)$, I get the general solution for $y$ as: $$(4)\ y=Ae^{4t}-6Be^{-3t}$$ Then it is straightforward to use the conditions stated to solve for $A$ and $B$. However, when solving a similar question, the textbook suggested a different method, by substituting $(3)$ into $(2)$ instead of $(1)$: $$\frac{dy}{dt}=6(Ae^{4t}+Be^{-3t})-2y$$ $$\frac{dy}{dt} + 2y =6Ae^{4t}+6Be^{-3t}$$ Then using the integrating factor method with the integrating factor as $e^{2t}$: $$ye^{2t}=\int 6Ae^{6t}+6Be^{-t} dt $$ $$ye^{2t}= Ae^{6t}-6Be^{-t}+C$$ $$y=Ae^{4t}-6Be^{-3t}+Ce^{-2t}$$ The solution is very similar to $(4)$, but with only two conditions, I can't find the value of all three constants. My question is, why do these two methods give different general solutions for $y$? Is there a reason that we shouldn't use the integrating factor method here, or a way to clearly show that $C=0$? Thanks so much for your help!

Answer 1

This has nothing to do with the method used to integrate. This has more to do with integration itself. In the first method you first differentiate and get a second order DE then you need to integrate twice to get $x(t)$. Two integration gives two constants. The you differentiate and used the second DE.

While in the second method you integrated the first DE twice this gives you two constants of integration then you integrated the second DE this gives you another constant of integration. Three integration performed three constants as expected. You can eliminate easily the extra constant of integration.

And to convince you you can solve the system with another method with integrating factor method and get only two constants of integration: First substract both DE and note that: $$y'-x'=-3(y-x)$$ $$((y-x)e^{3t})'=0$$ $$y(t)=x(t)+Ae^{-3t}$$ Now use this result to solve any of the original DE by integrating factor again. You only get two constants of integration since we only performed two integration.