Answer:
![y_g(t) = c_1*( 2t - 1 ) + c_2*e^(^-^2^t^) - e^(^-^2^t^)* [ t^3 + (3)/(4)t^2 + (3)/(4)t ]](https://img.qammunity.org/2021/formulas/mathematics/college/tmpskswh9w6iepgthv6ag12yr7p5fcq0fa.png)
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:
![u_1(t) = - \int {(F(t). y_2(t))/(W [ y_1(t) , y_2(t) ]) } \, dt \\\\u_2(t) = \int {(F(t). y_1(t))/(W [ y_1(t) , y_2(t) ]) } \, dt \\](https://img.qammunity.org/2021/formulas/mathematics/college/g2184b05d1djc7x2hhvwca58mw7v4jdzvo.png)
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:
![u_1 (t ) = - \int {(6t^2*e^(^-^2^t^) . ( e^-^2^t))/(-4t*e^(^-^2^t^)) } \, dt\\\\u_1 (t ) = (3)/(2) \int [ t*e^(^-^2^t^) ] \, dt\\\\u_1 (t ) = (3)/(2)* [ ( -(1)/(2) t*e^(^-^2^t^) - \int {( -(1)/(2)*e^(^-^2^t^) )} \, dt] \\\\u_1 (t ) = -e^(^-^2^t^)* [ ( (3)/(4) t + (3)/(8) )] \\\\](https://img.qammunity.org/2021/formulas/mathematics/college/oba4jbfbk2unjgbe49hklny34bhr6ckd8n.png)
- Similarly for the function u2(t):
![u_2 (t ) = \int {(6t^2*e^(^-^2^t^) . ( 2t-1))/(-4t*e^(^-^2^t^)) } \, dt\\\\u_2 (t ) = -(3)/(2) \int [2t^2 -t ] \, dt\\\\u_2 (t ) = -(3)/(2)* [(2)/(3)t^3 - (1)/(2)t^2 ] \\\\u_2 (t ) = t^2 [(3)/(4) - t ]](https://img.qammunity.org/2021/formulas/mathematics/college/wtpqld4sj4o67x1c4pviz86qd0icsagplg.png)
- We can now express the particular solution ( yp ) in the form expressed initially:
![y_p(t) = -e^(^-^2^t^)* [(3)/(2)t^2 + (3)/(4)t - (3)/(8) ] + e^(^-^2^t^)*[(3)/(4)t^2 - t^3 ]\\\\y_p(t) = -e^(^-^2^t^)* [t^3 + (3)/(4)t^2 + (3)/(4)t - (3)/(8) ] \\](https://img.qammunity.org/2021/formulas/mathematics/college/vud9ttmdcla6aehr98y4cvn3ykcp55r62b.png)
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: