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