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In Problems 1–8 use the method of undetermined coefficients to solve the given nonhomogeneous system.
4.
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Chapter 8 Solutions
First Course in Differential Equations (Instructor's)
- 3. 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_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_forwardExample 1.21 y" + 5y" + 12y' + 8y = 5sin2x + 10x? - 3x + 7 private solution yo =?arrow_forward
- 1. Use RK4-Systeml to solve each of the following for 0sIs1.Use h = 2-k with k = 5, 6, and 7, and compare results. (x" =x² - y +e x" = 2(e - x)/2 y" = x - y -e b. %3D a. x(0) = 0 x'(0) = 1 x (0) = 0 x'(0) = 0 y(0) = 1 y'(0) = -2arrow_forward02) a. Solve the following system by using Cramer's rule; 5x + 3y – 3z= -14, 3x+ z- 4y= 2; 7y+ z +x=6 b. Solve the differential equation; yy + 25x = 0arrow_forwardUse (1) in Section 8.4 X = eAtc (1) to find the general solution of the given system. 1 X' = 0. X(t) =arrow_forward
- 1. 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)(1)In three steps, describe how to write down the general solution of (1): Step 1: Step 2: Step 3:arrow_forward6. (2x +3y = 0 /2x x+2y =-1 9. (1 1 -X+-y 5 1, -x+y%3D10 4arrow_forwardQ. 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_forward
- Q.3// Use Tow Phase method to solve the Mathematical model (Stop at Table 1 in Stage Two) Max Z = X1 + 3xz+ 4x3+ 5 S.t. 4x1+ 3x3+ 2x3+x40arrow_forward10. Find the general solution of the system of differential equations 3 -2 -2 d. X = -3 -2 -6 X dt 3 10 1 + 2tet + 3t?et + 4t°et 3 1 -3 Hint: The characteristic polymomial of the coefficient matrix is -(A- 4)²(A- 3). Moreover (:) 2 1 Xp(t) = t²et +t³et +t'e3t -1 -1 -3 is a particular solution of the system.arrow_forward11. What is the general solution of* (2x – y)dx + (4x + y - 6)dy = 0 (2 +y – 3) = c(2x + y - 4)2 (x – y + 3)? = c(2æ + y – 4)3 Option 1 Option 2 (2 - y - 3) = c(2r - y- 4) (x+y - 3) = c(x + 2y – 4)?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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