Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Use the following method:

(a) by differentiating the first equation with respect to $t$ and eliminating $y$;
(b) by differentiating the second equation with respect to $t$ and eliminating $x$.

to find the general solution of the system.


Problem Number One: $$\frac{dx}{dt}=x+y$$ $$\frac{dy}{dt}=y$$ $$x'=x+y\space \space \rightarrow \space \space y=x'-x\space \space \rightarrow \space \space y'=x''-x'$$ $$y'=y$$ $$\text{So,}\space\space x''-x'=x'-x\space \space\rightarrow x''-2x'+x=0$$ $$\text{This is a second-order linear equation with constant coefficients so,}\space \space r^2-2r+r=0\rightarrow \space \space (r-1)^2=0 \space \space \text{where,}\space \space r=1$$

$$x=Ae^t+Bte^t$$ $$y=?$$


Problem Number Two:

$$\frac{dx}{dt}=x$$ $$\frac{dy}{dt}=y$$ $$x'=x$$ $$y'=y$$ How do I solve this? It seems even simpler than problem number one, but i'm stumped.

Part (b) of this problem says show that any second-order equation obtained from the system in part (a) is not equivalent to this system, in the sense that it has solutions that are not part of any solution of the system. Thus, although higher-order equations are equivalent to systems, the reverse is not true, and the systems are definitely more general.

share|improve this question
add comment

2 Answers

You get $\acute{x}=x+y$ and $x=Ae^{t}+Bte^{t}$, so we have $$y=\acute{x}-x$$ and $\acute{x}=x+Be^{t}$, thus we get $y=Be^{t}$.

share|improve this answer
add comment

For your second problem, $y$ does not depend on $x$ and $x$ does not depend on $y$. So solve the equations separately like you normally would.

For your first problem, you do have a relationship between $x$ and $y$: $y=x'-x$. You've already solved for $x$, so just plug in.

I assume there are tougher problems with the same instructions, where it is not so easy to eliminate one variable from an equation. If that's the case, it probably would not have been against the spirit of the problem to solve the second equation of problem 1 immediately to get $y=k_1e^t$, then solving $x'-x=k_1e^t$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.