Given yı (t) = t and y2(t) = tsatisfy the corresponding homogeneous equation oft'y" – 2y = 1 – 2t°, t > 0Then the general solution to the nonhomogeneous equation can be written asy(t) = c1y1(t) + ©2Y2(t) + Yp(t).Use variation of parameters to find a particular solution yp(t).Y,(t) =Tip: Before you use the formulaY29(t)Yı9(t)JW(n, y2)Yp = - Y1+ y2W (y1, Y2)the ODE should be in the form y'' + a(t)y' + b(t)y = g(t)

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Answered: Given yı (t) = t and y2(t) = t satisfy…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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A student correctly obtained the following characteristic equation: 2x² -4x2 +7x =0 What homogeneous linear equation is this student solving? O 2x³y" – 4x²y" +7xy=0 O 2y" – 4y"+7=0 O 2y" - 4y" +7y' = 0 o dy -2x³ – 4x² +7x 3D2X3 dx « Previous Not a 98
2. Suppose that y₁ is a solution for of the homogeneous SOLDE " + p(t)y' + q(t)y = 0. (a) Explain how to find a non constant function tv(t) such that y2(t) = v(t)y₁(t) is also a solution. (b) Explain why y₁ and y₂ are linearly independent.
a.) Form the complimentary solution to the homogeneous equation. y_c(t) = c_1 [ _ _] + c_2 [ _ _] b.) Construct a particular solution by assuming the form y_P(t) = e^(-3t)a and solving for the undetermined constant vector a. y_P(t) = [ _ _] c.) Form the general solution y(t) = y_c(t) + y_P(t) and impose the initial condition to obtain the solution of the initial value problem. [y_1(t) y_2(t)] = [ _ _]
Find the general solution by substitution or elimination Q6 A: = x -2 y +t dt —х-у+2t dt B: * = -y + 3e^-t x(0) = -1 = 2x + 3y У (0) — 1 %3D
Consider the followiing equation (t^2)y′′−6y =0, t > 0, check that y1 =t^3 and find y2 and write the solution of y(t) Choose your favorite method to find the second linearly independent solution (Abel's theorem or reduction or order). Choose the correct form of the general solution from the list below.
Solve the equation y' - graphically. Below, the graph in the (y, y') plane is shown. 10 (a) Find the equilibrium solutions and classify each as to its stability. (b) Determine where y(t) is increasing. (c) Determine where y(t) is concave up. (d) On the (t, y) plane, sketch the solutions with y(0) = -1, y(0) = 1 and y(0) = 6.
Show that the smoking free equilibrium E0 = (P0, S0, Qt0 ) = (1, 0, 0) and the smoking present equilibrium E* = (P *, S* , Q* t ), by setting the right-hand side of the equations in the model tozero.
A third-order homogeneous linear equation and three linearly independent solutions are given (you do not need to verify). Find a particular solution satisfying the given initial conditions. Show all work. y(³) — 5y" +8y′ − 4y = 0; y(0) = 1, y'(0) = 4, y″(0) = 0; y₁ = ex, y2 = e²x, y² = xe²x
The general solution of a nonhomogeneous linear equation has the following form: y= Yc + Yp What is yc? O General solution of the corresponding homogeneous equation. O Constant solution of the given equation. O Particular solution of the given equation. O Integrating factor. Next « Previous Not saved Submit Qu 曲
5. Okay, time for some mathematics. We're going to focus on the interactions between the hares and their main predators, the lynx. The tool we need is differential equations (last seen when we talked about the S-I-R disease model). We will let H(t) be the density of hares at time t, and L(t) the density of lynx at time t. Our first assumption is that, if there are no lynx around, the hares grow exponentially. Hares lead to more hares and faster growth. This gives us the differential equation (trust me): HP = aH dt where a is a positive constant. Second we assume that, if there are no hares around, the lynx starve and thus their numbers decay exponentially. This gives us: TP = -bL dt
Given y₁ (t) = t² and y2(t) = t¹ satisfy the corresponding homogeneous equation of t²y" - 2y = 2-t, t > 0, the general solution to the nonhomogeneous equation can be written as y(t) = yp(t) + C₁y₁(t) + c2y2(t). Use variation of parameters to find y, (t). Yp(t) Preview =
1. Please work out the solution for x(t). (1) Find the homogeneous solution: x(t) + 4x(t) + 3x(t) = 0. (Hint: Assume x(t) = ce^t) (2) Find the non-homogeneous solution: x(t) + 4x(t) + 3x(t) = 1. (3) Find the non-homogeneous solution: x(t) + 4x(t) + 3x(t) = sin (2t). 2. Please work out the homogeneous solution for the following 2nd-order ODES. (Hint: Assume x(t) = cet) (1) X(t) + 4x(t) + 4x(t) = 0. (2) X(t) + 4x(t) + 9x(t) = 0.

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