Higher Order Constant Coefficient Differential Equations: y”’+y’=0 and y””-3y”’+3y”-y’=0
Higher Order Constant Coefficient Differential Equations: y”’+y’=0 and y””-3y”’+3y”-y’=0

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Consider a system of two nonlinear third-order ordinary differential equations along with initial conditions (both should be randomly taken, i.e., of your own choice). An example of such a system is given below (do not consider it; your system and conditions should be different from the given):
y” = xy/2 + Vz, 2″ = Xyz” + yz, y(0) = 0.23, y'(0) = 0.14, y”(0) = 0.05, z(0) = 0.11, z'(0) = 0.19, z”(0) = 0.5.
Approximate the solution of the considered system at x = 0.5, with 0 < x < 1, h = 0.5, using the RK-method for systems after converting the considered system to a system of six first-order ODEs.

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02:37

We consider the following second order differential equation:d^2y/dt^2 – 2.25y.Write this equation as a system of two first order differential equations of the form:dy/dt = 0-4fwheref = [y, dy/dt]^Tand A is a 2×2 matrix. The matrix is equal to:A = [0, 1; -2.25, 0]The eigenvalues, A1 and A2, of the matrix A in ascending order (A1 < A2) are equal to:A1 = -1.5A2 = 1.5(i) Write the corresponding eigenvectors, U1 and U2, (U1 corresponds to A1 and U2 corresponds to A2) in their simplest form, such as their first component is v1 = (1, v2 = (1,U1 = [1, 1]U2 = [1, 1](iv) The general solution of this second order differential equation has the form:y(t) = C1 * exp(A1 * t) * U1 + C2 * exp(A2 * t) * U2where C1 and C2 are arbitrary complex numbers.The syntax for the exponential of x is exp(x). Please enter one answer only in each box (that is, don’t enter two answers in one box)(v) The solution satisfying this second order differential equation and the initial conditions y(0) = 0,…

08:01

We have not discussed methods by which systems of firstorder differential equations can be solved. Nevertheless, systems such as (2) can be solved with no knowledge other than how to solve a single linear first-order equation. Find a solution of (2) subject to the initial conditions \$x(0)=x_{0}, y(0)=0\$, \$z(0)=0\$

04:11

In this exercise, you will solve the initial value problem:

y” + 20y + 10y’ = y(0) = 5, y'(0) = -5. (1+x^2)

(1) Let C1 and C2 be arbitrary constants. The general solution to the related homogeneous differential equation y” + 20y’ + 10y = 0 is the function Yh(x) = C1Y1(x) + C2Y2(x) = C1 + C2.

NOTE: The order in which you enter the answers is important; that is, C1g(x) + C2f(x) â‰ C1f(x) + C2g(x).

(2) The particular solution yp(x) to the differential equation y” + 20y + 100y = 0 is of the form Yp(x) = Y1(x)u1(x) + Y2(x)u2(x), where u1(x) and u2(x) are functions.

(3) The most general solution to the non-homogeneous differential equation y” + 20y + 100y = (1+x^2) is:

y = Yh(x) + Yp(x) + âˆ«(1+x^2)dx.

Oops! There was an issue generating an instant solution