In Problems 11-16, use the variation of parameters formula (11) to find a general solution of the system
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Fundamentals of Differential Equations and Boundary Value Problems
- B1. Accountarrow_forwardDetermine two linearly independent solutions to x2y′′+xy′−(4+x)y =0, x > 0.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_forward
- If y, and y, are linearly independent solutions of ty" + 4y + tey = 0 and if W(₁₂)(1) = 4, find W(y₁ - y₂)(4). Round your answer to two decimal places. W(y₁+ y₂)(4) = iarrow_forward2. Find the solution y = f(x) to the equation x² + y²-2³ = 0 near each of the following points: (a) (5, 10) (b) (10,-30)arrow_forward1. Find steady states of the equation: 2xn (а) хр+1 Xn+1 K (b) Xn+1 where k1, k2 and K are constants. ki + k2/Xnarrow_forward
- Solve barrow_forwardThe goal of this problem is to show that is the quadratic approximation to f(x) = ex at the point a = 0. a). Consider the general form of a quadratic, namely Q(x) = bx2 + cx + d where b, c, d are constants. Compute Q′(x) and Q′′(x). b). Solve the system of equations Q(0) = f(0) and Q′(0) = f′(0) and Q′′(0) = f′′(0) for b, c, d.Deduce that ... as claimed.arrow_forwardx1(t) Let x(t) = be a solution to the system of differential equations: x2(t) x{(t) x½(t) -23 x1 (t) + 21 x2 (t) -28 x1(t) + 26 x2(t) 3 If æ(0) = , find æ(t). 5arrow_forward
- Use the variation of parameters formula to find a general solution of the system x'(t) = Ax(t) + f(t), where A and f(t) are given. - 16 4 [*][26] f(t)= 4 - 1 12+ 4t 1 A = - Xh(t) = Xp (t) = t1 Let X(t) = x₁(t) + xp (t), where x₁ (t) is the general solution corresponding to the homogeneous system, and xp (t) is a particular solution to the nonhomogeneous system. Find x₁ (t) and xp (t). (Type your answer as a single matrix.)arrow_forward6. (a) If u(t) and v(t) are solutions of the linear system (1), prove that for any constants a and b, w(t) = au(t) + bv(t) is a solution. (b) For A = [2] find solutions u(t) and v(t) of x = Ax such that every solution is a linear combination of u(t) and v(t).arrow_forwardThus far in the class we've only seen one kind of so-called "linear" difference equation models: the Malthusian ones of the form N(t+1)= XN (t) This is called linear because N(t+1) is a linear function of N(t) - in other words, if we view N(t+1) as "y" and N(t) as "r" then this is an equation of the form "y = mx" since À is a constant. Another more general type of "linear" difference equation is: N(t+1) = AN(t) + a where a is some constant. This is again a linear model because N(t+1) is a linear function of N(t), this time of the form "y=mx+b". This is sometimes called a "linear difference equation with a constant term". The next few problems have to do with linear difference equations with constant terms. 4. Consider the difference equation N(t + 1) = 0.5N(t) + 40 where we are given that N(0) = 100. a) Check that N(t) = 100* (0.5) is not a solution to this difference equation. b) Check that N(t) = 100* (0.5) +40 is not a solution to this difference equation. c) Check that N(t) = 20 *…arrow_forward
- Linear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage Learning