# Thread: find the general solution of the system...

1. ## find the general solution of the system...

ok so i was working this problem and got stuck, im not sure what to do with the imaginary numbers that im getting?

so here is the problem...

$\frac{dx}{dt}=6x-y , \frac{dy}{dt}=5x+2y$

i figured out that ...

$\lambda =4\pm i$

but im not sure where to go from here?

any help would be greatly appreciated

2. Divide the first equation by the second one: $\frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}=\frac{5x+2 y}{6x-y}$

that is $\frac{dy}{dx}=\frac{5x+2y}{6x-y}$

$y=vx$ gives $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Hence $v+x\frac{dv}{dx}=\frac{5x+2vx}{6x-vx}=\frac{5+2v}{6-v}$

So $x\frac{dv}{dx}=\frac{5+2v}{6-v}-v=\frac{5+2v-6v+v^2}{6-v}=\frac{v^2-4v+5}{6-v}$

Integrate this.

$\frac{v^2-4 v+5}{6-v}$

$=\frac{-v^2+4 v-5}{v-6}$

= $-v-\frac{-17}{(v-6)}-2$

= $-2\int dv-17\int \frac{1}{v-6} dv- \int v dv$

For the integrand $\frac{1}{v-6}$, substitute $u = v-6$
= $-1 7\int\frac{1}{u} du+ \int -2 dv- \int v dv$

= $-17 log(u)+ \int-2 dv- \int v dv$

= $-17 log(u)-\frac{v^2}{2}+ \int -2 dv$

= $-17 log(u)-\frac{v^2}{2}-2v$

= $-\frac{v^2}{2}-2v-17 log(v-6)$

Substitute back $v=\frac{y}{x}$ to get the final result.

3. Originally Posted by Sambit
Substitute back $v=\frac{y}{x}$ to get the final result.

All of this provides a prime integral F (x , y) of the autonomous system but not the solutions. That is, we determine the level sets F (x , y) = k where the orbits are contained. We can find the solutions for the system

$X'=AX \equiv\begin{bmatrix}{x'(t)}\\{y'(t)}\end{bmatrix} =\begin{bmatrix}{6}&{-1}\\{5}&{\;\;2}\end{bmatrix} \begin{bmatrix}{x(t)}\\{y(t)}\end{bmatrix}$

using the standard method

$X=e^{tA}\begin{bmatrix}{C_1}\\{C_2}\end{bmatrix}$

4. Originally Posted by Sambit
Divide the first equation by the second one: $\frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}=\frac{5x+2 y}{6x-y}$

that is $\frac{dy}{dx}=\frac{5x+2y}{6x-y}$

$y=vx$ gives $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Hence $v+x\frac{dv}{dx}=\frac{5x+2vx}{6x-vx}=\frac{5+2v}{6-v}$

So $x\frac{dv}{dx}=\frac{5+2v}{6-v}-v=\frac{5+2v-6v+v^2}{6-v}=\frac{v^2-4v+5}{6-v}$

Integrate this.

$\frac{v^2-4 v+5}{6-v}$

$=\frac{-v^2+4 v-5}{v-6}$

= $-v-\frac{-17}{(v-6)}-2$

= $-2\int dv-17\int \frac{1}{v-6} dv- \int v dv$

For the integrand $\frac{1}{v-6}$, substitute $u = v-6$
= $-1 7\int\frac{1}{u} du+ \int -2 dv- \int v dv$

= $-17 log(u)+ \int-2 dv- \int v dv$

= $-17 log(u)-\frac{v^2}{2}+ \int -2 dv$

= $-17 log(u)-\frac{v^2}{2}-2v$

= $-\frac{v^2}{2}-2v-17 log(v-6)$

Substitute back $v=\frac{y}{x}$ to get the final result.
That's sure is a lot of work and not very useful as Fernando points out, you want the solutions. One could also go the system route but what I would so is solve the first ODE for $y$, $y=6x - \dot{x}$ and substitute into the second giving

$\ddot{x} - 8 \dot{x} + 17x = 0$

Characteristic equation

$m^2-8m + 17 = 0$ so $m = 4 \pm i$

soln: $x =c_1e^{4t} \cos t + c_2 e^{4t} \sin t$

Then sub. this into $y=6x - \dot{x}$

5. Originally Posted by FernandoRevilla
All of this provides a prime integral F (x , y) of the autonomous system but not the solutions. That is, we determine the level sets F (x , y) = k where the orbits are contained. We can find the solutions for the system

$X'=AX \equiv\begin{bmatrix}{x'(t)}\\{y'(t)}\end{bmatrix} =\begin{bmatrix}{6}&{-1}\\{5}&{\;\;2}\end{bmatrix} \begin{bmatrix}{x(t)}\\{y(t)}\end{bmatrix}$

using the standard method

$X=e^{tA}\begin{bmatrix}{C_1}\\{C_2}\end{bmatrix}$
ok so i will go with the standard method and have another go at the problem. thanks you guys!

6. Originally Posted by Danny
That's sure is a lot of work and not very useful as Fernando points out, you want the solutions. One could also go the system route but what I would so is solve the first ODE for $y$, $y=6x - \dot{x}$ and substitute into the second giving

$\ddot{x} - 8 \dot{x} + 17x = 0$

Characteristic equation

$m^2-8m + 17 = 0$ so $m = 4 \pm i$

soln: $x =c_1e^{4t} \cos t + c_2 e^{4t} \sin t$

Then sub. this into $y=6x - \dot{x}$
thanks Danny! I was looking through another textbook and saw that similiar form that you gave... the author of the text was saying something to the effect of $\alpha \pm \beta i$ im going to read through it again and redo this problem, it doesnt seem bad after reading through the explanations posted! This forum is the reason why im passing Diffy Q!