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In Problems 13–32 use variation of parameters to solve the given nonhomogeneous system.
21.
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Chapter 8 Solutions
FIRST COURSE IN DIFF.EQ.-WEBASSIGN
- 1. Find the general solution of y"" - 2y" - y' + 2y = 0arrow_forwardSolve 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_forward7. A scientist places two strains of bacteria, X and Y, in a petri dish. Initially, there are 400 of X and 500 of Y. The two bacteria compete for food and space but do not feed on each other. If x = x(t) and y = y(t) are the numbers of strains at time t days, the growth rates of the two populations are given by the system x' = 1.2x – 0.2y, y' = -0.2x + 1.5y Determine what happens to these two populations by solving the system of differential equations.arrow_forward
- 1. Find the solution to the initial value problem 4x3 + 1 2у — 6 y(1) = 2. A. y = 3 – Vxª + x – 1 B. y = 2+ Vx³ + x – 2 C. y = 1+ Vx4 + x – 1 D. y = 4 – V4x³ + x – 1 E. y = V4³ + x – 1arrow_forwardExample 2: Find the particular solution of (x – y)dx + (3x+y)dy = 0 when x = 2 & y = -1.arrow_forwardSuppose 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
- Q. No. 11 The solution of the DE 3ry" + y/ – y = 0 (a) yı = rš[1 – {x +²+...], y2 = 1+x – 20² + ... (b) yı = a3[1 – r +a² + ...], y2 = 1+ 2x – 2x² + ... (c) yı = xš[1 – x + a² + ...], y2 =1+ 2x – 2x3 + ... (d) yı = [1 – x + x² + ...], y2 = 1+ 2x – 2x2 +... solve this and tick the correct optionarrow_forward1. Show that y = ze* + e-2* is a solution of y' + 2y = 2e*. 3arrow_forwardExample 10.34. Solve the equation y" = x + y with the boundary conditions y(0) = y(1) = 0.arrow_forward
- 5. Find a general solution of the linear system. x' = 2x + y ly' = x + 2y – e2tarrow_forwardProblem 2. Consider the equation: x?y"(x) – xy' +y = 0. Given that yı(x) = x is a solution of this equation. Use the method of reduction of order, find the second solution y2(x) of the equation so that y1 and y2 are linearly independent. (Hint: y2(x) should be given in the form y2(x) = u(x)y1(x). Substitute it into the equation to find u(x).) %3Darrow_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_forward
- Discrete Mathematics and Its Applications ( 8th I...MathISBN:9781259676512Author:Kenneth H RosenPublisher:McGraw-Hill EducationMathematics for Elementary Teachers with Activiti...MathISBN:9780134392790Author:Beckmann, SybillaPublisher:PEARSON
- Thinking Mathematically (7th Edition)MathISBN:9780134683713Author:Robert F. BlitzerPublisher:PEARSONDiscrete Mathematics With ApplicationsMathISBN:9781337694193Author:EPP, Susanna S.Publisher:Cengage Learning,Pathways To Math Literacy (looseleaf)MathISBN:9781259985607Author:David Sobecki Professor, Brian A. MercerPublisher:McGraw-Hill Education
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