Solving a system of ODEs with substitution/elimination

[Lahooty]

Homework Statement

Use substitution/elimination to find the general solution.

dx/dt = 3x-y+2t^2

dy/dt = 4x-2y-8t^2

Homework Equations

I'm practically clueless on how to solve this problem using substitution/elimination, I'm pretty sure the way I'm doing it is completely wrong. If anyone can show me the correct approach or can show an example it would be a tremendous help.

The Attempt at a Solution

dx/dt = 3x - y +2t^2

dx = (3x -y +2t^2)dt

x = 3xt - yt + 2/3t^3

x(1-3t) = - yt + 2/3t^3

x = (-yt + 2/3t^3)/(1-3t)

I know this is not the correct way but I can't find anything online, so if someone does know what to do please reply.

Thanks

 

[pasmith]

The Attempt at a Solution

dx/dt = 3x - y +2t^2

dx = (3x -y +2t^2)dt

x = 3xt - yt + 2/3t^3

x is a function of t. If you integrate it, you get $\int x(t)\,\mathrm{d}t$, not xt. Similarly for y.

Homework Statement

Use substitution/elimination to find the general solution.

dx/dt = 3x-y+2t^2

dy/dt = 4x-2y-8t^2

We want to eliminate either x or y and its derivatives to get a second order ODE for whichever variable we're left with.

Here it seems easiest to eliminate y, since the first equation tells us that
$$y = 3x - \frac{\mathrm{d}x}{\mathrm{d}t} + 2t^2$$
Differentiating this gives
$$\frac{\mathrm{d}y}{\mathrm{d}t} = 3\frac{\mathrm{d}x}{\mathrm{d}t} <br /> - \frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 4t$$
You can now substitute these expressions for y and dy/dt into the second equation.

 

[Lahooty]

Thanks for the response and help. Alright so here is what I did:

y = 3x- x' + 2t^2;

(3x- x' + 2t^2)' = 4x -2(3x- x' + 2t^2) + 2t^2;

3x' - x'' + 2t^2' = 4x -6x + 2x' -4t^2 + 2t^2

x'' - x' -2x = 2t^2' - 2t^2

r^2 - r - 2 = 0

r = 2, r = -1

I don't know how to proceed from this point on.

 

[pasmith]

Thanks for the response and help. Alright so here is what I did:

y = 3x- x' + 2t^2;

(3x- x' + 2t^2)' = 4x -2(3x- x' + 2t^2) + 2t^2;

The second equation is $y' = 4x - 2y - 8t^2$, so you should have
$$(3x- x' + 2t^2)' = 4x -2(3x- x' + 2t^2) - 8t^2$$
which yields
$$3x' - x'' + 4t = 4x - 6x + 2x' - 4t^2 - 8t^2$$
(because $(2t^2)' = 2(2t) = 4t$) so that
$$x'' - x' - 2x = 4t + 12t^2$$

r^2 - r - 2 = 0

r = 2, r = -1

I don't know how to proceed from this point on.

You need to find a solution of
$$x'' - x' - 2x = 4t + 12t^2$$
and add to it the general solution of
$$x'' - x' - 2x = 0$$
The second part you should know. For the first part, the form of the right hand side suggests taking $f(t) = at^2 + bt + c$ with constants a, b and c, which you should choose so that $f'' - f' - 2f = 4t + 12t^2$.

That gives you $x(t)$, and you can then find $y(t)$ from $y = 3x- x' + 2t^2$.