Explanation The given differential equation is y ″ + ω 2 y = sin ( n t ) . Consider the corresponding homogeneous differential equation y ″ + ω 2 y = 0 . The characteristic equation corresponding to y ″ + ω 2 y = 0 is m 2 + ω 2 = 0 . Compute the roots of m 2 + ω 2 = 0 as shown below. m 2 + ω 2 = 0 m 2 = − ω 2 m = ± i ω The roots are m = + i ω and − i ω . Since the characteristic equation has complex conjugate roots, the complementary function is given by y h = c 1 cos ω t + c 2 sin ω t . Compute the particular integral of the nonhomogeneous differential equation y ″ + ω 2 y = sin ( n t ) as shown below. The undetermined coefficient set corresponding to the right hand side function sin ( n t ) is { sin n t , cos n t } . Consider the linear combination y = d 1 sin n t + d 2 cos n t of sin n t and cos n t . Compute the first and the second derivatives of the equation y = d 1 sin n t + d 2 cos n t . y = d 1 sin n t + d 2 cos n t y ′ = d 1 n ( cos n t ) + d 2 n ( − sin n t ) y ″ = d 1 n 2 ( − sin n t ) + d 2 n 2 ( − cos n t ) y ″ = − d 1 n 2 sin n t − d 2 n 2 cos n t Substitute for y , y ′ and y ″ in the differential equation y ″ + ω 2 y = sin ( n t ) . y ″ + ω 2 y = sin ( n t ) ( − d 1 n 2 sin n t − d 2 n 2 cos n t ) + ω 2 ( d 1 sin n t + d 2 cos n t ) = sin ( n t ) ( − d 1 n 2 + ω 2 d 1 ) sin n t + ( − d 2 n 2 + ω 2 d 2 ) cos n t = sin ( n t ) d 1 ( ω 2 − n 2 ) sin n t + d 2 ( ω 2 − n 2 ) cos n t = sin n t Compare the coefficients of sin n t and cos n t on both sides of the above equation. d 1 ( ω 2 − n 2 ) = 1 d 1 = 1 ω 2 − n 2 And, d 2 ( ω 2 − n 2 ) = 0 d 2 = 0 That is, the value of d 1 = 1 ω 2 − n 2 , d 2 = 0 . Substitute d 1 = 1 ω 2 − n 2 , d 2 = 0 in y = d 1 sin n t + d 2 cos n t . Therefore, y p = sin n t ω 2 − n 2 is the particular integral of the differential equation y ″ + ω 2 y = sin ( n t ) . The general solution of y ″ + ω 2 y = sin ( n t ) is given by y = y h + y p . Hence, y = c 1 cos ω t + c 2 sin ω t + sin n t ω 2 − n 2 is the general solution of the differential equation y ″ + ω 2 y = sin ( n t ) . Use the given initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 0 to evaluate c 1 and c 2 . Substitute the values t = 0 and y ( 0 ) = 0 in the general solution y = c 1 cos ω t + c 2 sin ω t + sin n t ω 2 − n 2 as follows. y ( 0 ) = 0 c 1 cos ω ( 0 ) + c 2 sin ω ( 0 ) + sin n ( 0 ) ω 2 − n 2 = 0 c 1 + 0 + 0 = 0 c 1 = 0 Substitute the value c 1 = 0 in the general solution y = c 1 cos ω t + c 2 sin ω t + sin n t ω 2 − n 2 . Therefore, the general solution becomes y = c 2 sin ω t + sin n t ω 2 − n 2 . Compute the derivative of above y as follows. y ( t ) = c 2 sin ω t + sin n t ω 2 − n 2 y ′ ( t ) = ω c 2 cos ω t + n cos n t ω 2 − n 2 Substitute t = 0 and y ′ ( 0 ) = 0 in the general solution y = c 2 sin ω t + sin n t ω 2 − n 2 as follows. y ′ ( 0 ) = 0 ω c 2 cos ω ( 0 ) + n cos n ( 0 ) ω 2 − n 2 = 0 ω c 2 = − n ω 2 − n 2 c 2 = − n ω ( ω 2 − n 2 ) Therefore, the arbitrary constant c 2 can take any value. Substitute the value of c 2 in y = c 2 sin ω t + sin n t ω 2 − n 2 as follows. y = − n ω ( ω 2 − n 2 ) sin ω t + sin n t ω 2 − n 2 = ω sin n t − n sin ω t ω ( ω 2 − n 2 ) [ ∵ ω 2 ≠ n 2 ] Hence, the solution of the initial value problem y ″ + ω 2 y = sin ( n t ) , with the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 0 is y = ω sin n t − n sin ω t ω ( ω 2 − n 2 ) _ for ω 2 ≠ n 2