elimination

[logistic_guy]

Use systematic elimination to solve the given system.

$\displaystyle \frac{dx}{dt} + \frac{dy}{dt} = 2x + 2y + 1$

$\displaystyle \frac{dx}{dt} + 2\frac{dy}{dt} = y + 3$

 

[khansaheb]

Use systematic elimination to solve the given system.

$\displaystyle \frac{dx}{dt} + \frac{dy}{dt} = 2x + 2y + 1$

$\displaystyle \frac{dx}{dt} + 2\frac{dy}{dt} = y + 3$

show us your effort/s to solve this problem.

 

[logistic_guy]

First step we change the differential operator from $\displaystyle \frac{d}{dt}$ to $\displaystyle D$.

$\displaystyle Dx + Dy = 2x + 2y + 1$

$\displaystyle Dx + 2Dy = y + 3$

Second step we simplify by combining like-term variables.

$\displaystyle (D - 2)x + (D - 2)y = 1$

$\displaystyle Dx + (2D - 1)y = 3$

 

[logistic_guy]

To get rid of $\displaystyle x$, we multiply (not normal multiplication) the first equation by $\displaystyle D$ and the second equation by $\displaystyle -(D - 2)$. Then, we add the first and second equations together.

$\displaystyle D(D - 2)x + D(D - 2)y = D(1)$

$\displaystyle -(D - 2)Dx - (D - 2)(2D - 1)y = -(D - 2)(3)$

After addition, we get:

$\displaystyle D(D - 2)y - (D - 2)(2D - 1)y = D(1) - (D - 2)(3)$

After we simplify we get:

$\displaystyle (D^2 - 2D)y - (2D^2 - D - 4D + 2)y = D(1) - D(3) + 6$

Now look at the right side and be careful as $\displaystyle D(1) = \frac{d(1)}{dt} = 0$. (Derivative of a constant is zero.)

Then,

$\displaystyle (D^2 - 2D)y - (2D^2 - D - 4D + 2)y = 0 - 0 + 6$

Let us simplify further.

$\displaystyle (D^2 - 2D)y - (2D^2 - 5D + 2)y = 6$

$\displaystyle (-D^2 + 3D - 2)y = 6$

$\displaystyle (D^2 - 3D + 2)y = -6$

This is just a nonhomogeneous ordinary second order differential equation. And you can write it as:

$\displaystyle \frac{d^2y}{dt^2} - 3\frac{dy}{dt} + 2y = -6$, if you want.

We already know how to solve this type of differential equations. We start by solving the homogeneous version of that equation which is:

$\displaystyle (D^2 - 3D + 2)y = 0$

And that will be our mission in the next post.

 

[logistic_guy]

Last time we got this homogeneous differential equation: $\displaystyle (D^2 - 3D + 2)y = 0$.

Let us find the roots of its characteristic equation.

$\displaystyle r = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)} = \frac{3 \pm 1}{2}$

This means,

$\displaystyle r_1 = 2$
$\displaystyle r_2 = 1$

And the complementary solution of that differential equation is:

$\displaystyle y_c(t) = c_1e^{2t} + c_2e^{t}$

In the next post, we will try to find a particular solution of the form:

$\displaystyle y_p(t) = A$

where $\displaystyle A$ is a constant.

 

[logistic_guy]

We have the nonhomogeneous differential equation $\displaystyle (D^2 - 3D + 2)y = -6$ and we want to find a particular solution of the form $\displaystyle y_p(t) = A$.

We have:

$\displaystyle D y_p = 0$
$\displaystyle D^2 y_p = 0$

Then,

$\displaystyle (0 - 3(0) + 2)A = -6$

$\displaystyle A = -3$

And the general solution of $\displaystyle y(t)$ is:

$\displaystyle y(t) = y_c(t) + y_p(t) = c_1e^{2t} + c_2e^{t} - 3$

 

[logistic_guy]

Let us go back to the system again.

$\displaystyle (D - 2)x + (D - 2)y = 1$

$\displaystyle Dx + (2D - 1)y = 3$

Now we want to get rid of $\displaystyle y$, so we multiply (not normal multiplication) the first equation by $\displaystyle (2D - 1)$ and the second equation by $\displaystyle -(D - 2)$, then we add them together.

$\displaystyle (2D - 1)(D - 2)x + (2D - 1)(D - 2)y = (2D - 1)(1)$

$\displaystyle -(D - 2)Dx - (D - 2)(2D - 1)y = -(D - 2)(3)$

After addition, we get:

$\displaystyle (2D - 1)(D - 2)x - (D - 2)Dx = (2D - 1)(1) - (D - 2)(3)$

In the next post, we will try to simplify it further, and solve it.

 

[logistic_guy]

Last time we got this:

$\displaystyle (2D - 1)(D - 2)x - (D - 2)Dx = (2D - 1)(1) - (D - 2)(3)$

Let's simplify it.

$\displaystyle (2D^2 - 4D - D + 2)x - (D^2 - 2D)x = 2D(1) - 1 - D(3) + 6$

$\displaystyle (2D^2 - 5D + 2)x - (D^2 - 2D)x = 2(0) - 1 - 0 + 6$

$\displaystyle (D^2 - 3D + 2)x = 5$

This is just a nonhomogeneous ordinary second order differential equation. And we know how to solve it