Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linearly independent solutions to the corresponding homogeneous equation for t > 0. ty" + (2t - 1 )y' - 2y = 6t^2 e^-2t​; y1 = 22t −​1, y2 = e^-2t

Step-by-step explanation:

Solution:-

- Given is the 2nd order linear ODE as follows:

- The complementary two independent solution to the homogeneous 2nd order linear ODE are given as follows:

- The particular solution ( yp ) to the non-homogeneous 2nd order linear ODE is expressed as:

Where,

              are linearly independent functions of parameter ( t )

- To determine [   ], we will employ the use of wronskian ( W ).

- The functions [ ] are defined as:

Where,

      F(t): Non-homogeneous part of the ODE

      W [ y1(t) , y2(t) ]: the wronskian of independent complementary solutions

- To compute the wronskian W [ y1(t) , y2(t) ] we will follow the procedure to find the determinant of the matrix below:

- Now we will evaluate function. Using the relation given for u1(t) we have:

- Similarly for the function u2(t):

- We can now express the particular solution ( yp ) in the form expressed initially:

Where the term: 3/8 e^(-2t) is common to both complementary and particular solution; hence, dependent term is excluded from general solution.

- The general solution is the superposition of complementary and particular solution as follows:

i believe the answer to be 48

step-by-step explanation:

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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