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In Problems 13–32 use variation of parameters to solve the given nonhomogeneous system.
18.
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A First Course in Differential Equations with Modeling Applications (MindTap Course List)
- Solve each second-order IVP. 1. y" + 2y – 15y = 0, y(0) = 2, /(0) = -6 . 2. y" + 6y' + 13y = 0, y(0) = -1, y (0) = 5 3. y" + 2y +y = 0, y(0) = 3, y'(0) = –1arrow_forwardExample 2: Find the particular solution of (x – y)dx + (3x+y)dy = 0 when x = 2 & y = -1.arrow_forward3. 2хydx - (3xу + 2y?)dy %3D0 o (x - 2y)*(2x +y) = c (х — у)"(х + у) %3 с (х + 2y) (2х- у)* %3 с (x – 2y)* = c(2x + y)arrow_forward
- Suppose we are given y1(x) and y2(x) (with y1 # y2), which are two different solutions of a nonhomogeneous equation y" + p(x)y + q(x)y = g(x) In three steps, describe how to write down the general solution of (1): (1) Step 1: Step 2: Step 3:arrow_forward.1) The general solution of the equation y" +y =-5e-* , is: o y(x) = Ce-x + ez (Acosx + Bsinx) -- O y(x) = Ce-* + ez (Acosx + Bsinx) -xe-x Oy(x) = Ce-* + ez(Acosx + Bsinx) +xe-* %3D ) y(x) = Ce-* + ez (Acosx + Bsinx) +e-* %3Darrow_forward1. Show that y = ze* + e-2* is a solution of y' + 2y = 2e*. 3arrow_forward
- In Problems 1–8 use the method of undetermined coeffi cients to solve the given system. dx 1. = 2x + 3y – 7 dt dy 2у + 5 = -x - dt dx 2. dt 5х + 9у + 2 dy — —х + 11у +6 dt -G )x+ (,) 3 3. X' = 3 t (4t + 9er\ 4. X' = 4 -4 х+ 1 -t + e6t 3 X + le' 10 5. X' = sin t 5 X + 1 6. X' 1 -2 cos 7. X' = 2 3 X +| -1 \0 0 5, 2 8. X' = 5 0 ]X + - 10 5 0 0/ 40/arrow_forward1. The Lotka-Volterra or predator-prey equations dU = aU – UV, dt (1) AP = eyUV – BV. dt (2) have two fixed points (U., V.) = (0,0), (U., V.) = (- :). The trivial fixed point (0,0) is unstable since the prey population grows exponentially if it is initially small. Investigate the stability of the second fixed point (U..V.) = 6:27 PM 3/3/2021 近arrow_forwardDetermine the solution of the initial value problem 1. 5y" +y' 2. y" + 4y' + 5y = 35e4*, у (0) — 0, у'(0) 3 —10 У (0) 3D — 3, у'(0) — 1 = -6x,arrow_forward
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