The method of reduction of order (Section 3.4) can also be used for the nonhomogeneous equation y" +p(t)y' + q(t)y = g(). (38) provided one solution y of the corresponding homogeneous equation is known. Let y= ()y() and show that y satisfies equation 38 if v is a solution of yı(" + (2y,) + p(t)y = g(0). (39) Equation 39 is a first-order linear differential equation for v. By solving equation 39 for v, integrating the result to find v, and then multiplying by y1(O. you can find the general solution of equation 38. This method simultaneously finds both the second homogeneous solution and a particular solution. Use the method above to solve the differential equation ty" – (1+t)y' +y = 5t2e24, t > 0, y (t) = 1+ t. Use C1, C2, ... for the constants of integration.

This is the first part of a two-part problem. Let P=[-: 1 5₁(t) = [(41) 5₂(t) = - sin(4t). a. Show that y₁ (t) is a solution to the system ÿ' = Pÿ by evaluating derivatives and the matrix product y(t) = = 0 [1] -4 Enter your answers in terms of the variable t. -4 sin(4t) -4 cos(4t)] ÿ₁ (t) [181-18] b. Show that y₂ (t) is a solution to the system ÿ' = Pÿ by evaluating derivatives and the matrix product Enter your answers in terms of the variable t. 04] 32(t) = [-28]|2(t) 181-181

As a specific example we consider the non-homogeneous problem y" + 6y' + 8y = (1) The general solution of the homogeneous problem (called the complementary solution, Yc Ay₁ + By2 ) is given in terms of a pair of linearly independent solutions, Y₁, Y2. Here A and B are arbitrary constants. Find a fundamental set for y” + 6y' + 8y = 0 and enter your results as a comma separated list e^(-2x),e^(-4x) (2) Choose particula Y₁, Y2 in the fundamental set such that the W(x) is positive. Then W (x) 2e^(-6x) U 1 = If your W (x) is negative, then you need to switch your choise of y₁ and y2 and recalculate your W(x). The y₁= e^(-2x) ՂԱԶ √= With this appropriate order we are now ready to apply the method of variation of parameters. -Y₂(x) f(x) W(x) y = Yc + Yp and the y₂2 Y₁(x)f(x) W(x) S [21 dx -cos(e^(2x)) = dx = And combining these results we arrive at = S S e^(-4x) = 2e^(4x)sin(e^(2x)) dx 4 sin(e²x) dx = = = Ур (3) Finally, using A and B for the arbitrary constants in yc, the general solution…

Solutions, if they exist, of equations of the form an(x)y(m) (x) + an-1(x)y(n-1)(x)+ ... + a1 (x)y'(x) + ao(x)y(x) = f(x) are completely described by n linearly independent functions. Please, stop posting the same copied solution from chegg. I need to prove this is false using the EXISTENCE AND UNIQUENESS THEOREM. I know the answer, please just explain the E&U aspect of it.