In Problems 21–30 find the general solution of the given system.
27.
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A First Course in Differential Equations with Modeling Applications (MindTap Course List)
- 4. (S.10). Use Gaussian elimination with backward substitution to solve the following linear system: 2.r1 + 12 – 13 = 5, 1 + 12 – 3r3 = -9, -I1 + 12 +2r3 = 9;arrow_forwardExample 1.21 y" + 5y" + 12y' + 8y = 5sin2x + 10x? - 3x + 7 private solution yo =?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
- 5. Find the general solution of the given system. X' = [; x. Х.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_forward2. Find the general solution of y (4) + 2y" + y = 0arrow_forward
- Determine the solution of (2x - 3y + 2)dx + (2x - 3y + 1)dy = 0. a. 10x + 10y + ln(10x - 15y + 8)2 = c b. 10x + 10y + ln(10x - 15y - 8)2 = c c. 10x - 10y + ln(10x + 15y + 8)2 = c d. 10x - 10y - ln(10x + 15y + 8)2 = carrow_forward4. Suppose the following functions are a general solution of: y(4) + a3y" +a2y" + a1y' + a0y = 0arrow_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
- 9. P = 15 -4 -7 2e31 – 8e- -4e31 + 2e- ž(1) = | 3e3t – 20e- -6e31 + 5et Show that x1 (t) is a solution to the system x = Px by evaluating derivatives and the matrix product -4 ž(1) = | 15 -7 Enter your answers in terms of the variable t. Show that x2(t) is a solution to the system x' = Px by evaluating derivatives and the matrix product 9. 3(1) = | 15 -4 X2(t) -7 Enter your answers in terms of the variable t.arrow_forwardQuestion. Solve: x" -x + 5y² = + 4" - 4y - 2x² = -2 with x (0) = y(0) = x²(0) = y₁ (0) = 0 Let & [ X (²) } = U (6) & & Gy (+) 2 = V(c) Evaluate U(2) Evaluate (²) Evaluate x(1) Evaluate y(1)arrow_forward2. x + y – z = 2 2x + 2y - 2z = 6 5x + y – 3z = 8arrow_forward
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