13. Suppose that one solution y1(x) of a homogenous Choose the answer below for y, that solves the differential second-order linear differential equation is known (on an interval I where p and q are continuous functions). equation and is linearly independent from y1. 0= (x)b + ,K(x)d + ,,K The method of reduction of order consists of substituting O A. e 2x О в. е 4x Oc. O D. ex y2(x)=v(x)y1(x) into the differential equation above and Oc. %3D attempting to determine the function v(x) so that y2(x) is a OE. OE. OF. second linearly independent solution of the differential equation. It can be shown that this substitution leads to the following equation, which is a separable equation that is readily solved for the derivative v'(x) of v(x). Integration of v'(x) then gives the desired (nonconstant) function v(x). yıv"+ (2y1' + py1) v' = 0 = ,^ (T ) A differential equation and one solution y1 is given below. Use the method of reduction of order to find a second linearly independent solution y2. x²y" - xy' – 8y = 0; (x > 0); y1(x) = xª 14. Suppose that one solution y1(x) of a homogenous Choose the answer below for y2 that solves the differential second-order linear differential equation is known (on an interval I where p and q are continuous functions). equation and is linearly independent from y1. y" + p(x)y' + q(x)y = 0 O A. X-2 О в. е 2х The method of reduction of order consists of substituting O C. xex O D. xe -X y2(X) = v(x)y1 (X) into the differential equation above and O E. X+2 O F. e-X %3D attempting to determine the function v(x) so that y2(x) is a second linearly independent solution of the differential equation. It can be shown that this substitution leads to the following equation, which is a separable equation that is readily solved for the derivative v'(x) of v(x). Integration of v'(X) then gives the desired (nonconstant) function v(x). Yıv"+ (2y1'+py1)v'=0 A differential equation and one solution y1 is given below. Use the method of reduction of order to find a second linearly independent solution y2. (x + 1)y" – (x + 2)y' +y = 0; (x > - 1); y1(x)= e× |