Respuesta :
Answer:
[tex]y_g(t) = c_1*( 2t - 1 ) + c_2*e^(^-^2^t^) - e^(^-^2^t^)* [ t^3 + \frac{3}{4}t^2 + \frac{3}{4}t ][/tex]
Step-by-step explanation:
Solution:-
- Given is the 2nd order linear ODE as follows:
[tex]ty'' + ( 2t - 1 )*y' - 2y = 6t^2 . e^(^-^2^t^)[/tex]
- The complementary two independent solution to the homogeneous 2nd order linear ODE are given as follows:
[tex]y_1(t) = 2t - 1\\\\y_2 (t ) = e^-^2^t[/tex]
- The particular solution ( yp ) to the non-homogeneous 2nd order linear ODE is expressed as:
[tex]y_p(t) = u_1(t)*y_1(t) + u_2(t)*y_2(t)[/tex]
Where,
[tex]u_1(t) , u_2(t)[/tex] are linearly independent functions of parameter ( t )
- To determine [ [tex]u_1(t) , u_2(t)[/tex] ], we will employ the use of wronskian ( W ).
- The functions [[tex]u_1(t) , u_2(t)[/tex] ] are defined as:
[tex]u_1(t) = - \int {\frac{F(t). y_2(t)}{W [ y_1(t) , y_2(t) ]} } \, dt \\\\u_2(t) = \int {\frac{F(t). y_1(t)}{W [ y_1(t) , y_2(t) ]} } \, dt \\[/tex]
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:
[tex]W [ y_1 ( t ) , y_2(t) ] = | \left[\begin{array}{cc}y_1(t)&y_2(t)\\y'_1(t)&y'_2(t)\end{array}\right] |[/tex]
[tex]W [ (2t-1) , (e^-^2^t) ] = | \left[\begin{array}{cc}2t - 1&e^-^2^t\\2&-2e^-^2^t\end{array}\right] |\\\\W [ (2t-1) , (e^-^2^t) ]= [ (2t - 1 ) * (-2e^-^2^t) - ( e^-^2^t ) * (2 ) ]\\\\W [ (2t-1) , (e^-^2^t) ] = [ -4t*e^-^2^t ]\\[/tex]
- Now we will evaluate function. Using the relation given for u1(t) we have:
[tex]u_1 (t ) = - \int {\frac{6t^2*e^(^-^2^t^) . ( e^-^2^t)}{-4t*e^(^-^2^t^)} } \, dt\\\\u_1 (t ) = \frac{3}{2} \int [ t*e^(^-^2^t^) ] \, dt\\\\u_1 (t ) = \frac{3}{2}* [ ( -\frac{1}{2} t*e^(^-^2^t^) - \int {( -\frac{1}{2}*e^(^-^2^t^) )} \, dt] \\\\u_1 (t ) = -e^(^-^2^t^)* [ ( \frac{3}{4} t + \frac{3}{8} )] \\\\[/tex]
- Similarly for the function u2(t):
[tex]u_2 (t ) = \int {\frac{6t^2*e^(^-^2^t^) . ( 2t-1)}{-4t*e^(^-^2^t^)} } \, dt\\\\u_2 (t ) = -\frac{3}{2} \int [2t^2 -t ] \, dt\\\\u_2 (t ) = -\frac{3}{2}* [\frac{2}{3}t^3 - \frac{1}{2}t^2 ] \\\\u_2 (t ) = t^2 [\frac{3}{4} - t ][/tex]
- We can now express the particular solution ( yp ) in the form expressed initially:
[tex]y_p(t) = -e^(^-^2^t^)* [\frac{3}{2}t^2 + \frac{3}{4}t - \frac{3}{8} ] + e^(^-^2^t^)*[\frac{3}{4}t^2 - t^3 ]\\\\y_p(t) = -e^(^-^2^t^)* [t^3 + \frac{3}{4}t^2 + \frac{3}{4}t - \frac{3}{8} ] \\[/tex]
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:
[tex]y_g(t) = y_c(t) + y_p(t)\\\\y_g(t) = c_1*( 2t - 1 ) + c_2*e^(^-^2^t^) - e^(^-^2^t^)* [ t^3 + \frac{3}{4}t^2 + \frac{3}{4}t ][/tex]
