Substitution/Elimination to find General Solution | System of ODE's

[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&#039; = 4x - 2y - 8t^2, so you should have
<br /> (3x- x&#039; + 2t^2)&#039; = 4x -2(3x- x&#039; + 2t^2) - 8t^2<br />
which yields
<br /> 3x&#039; - x&#039;&#039; + 4t = 4x - 6x + 2x&#039; - 4t^2 - 8t^2<br />
(because (2t^2)&#039; = 2(2t) = 4t) so that
<br /> x&#039;&#039; - x&#039; - 2x = 4t + 12t^2<br />

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
<br /> x&#039;&#039; - x&#039; - 2x = 4t + 12t^2<br />
and add to it the general solution of
<br /> x&#039;&#039; - x&#039; - 2x = 0<br />
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&#039;&#039; - f&#039; - 2f = 4t + 12t^2.

That gives you x(t), and you can then find y(t) from y = 3x- x&#039; + 2t^2.