# I example find a particular solution to the equation

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I Example Find a particular solution to the equation y 00 + 4 y = (1 + x ) cos(2 x ) Solution In this case we have p ( x ) = 1+ t and f ( x ) = cos(2 x ) where f ( x ) is the real part of the complex-valued function ˆ f ( x ) = e 2 ix . Then we will apply the method of Case 2 to find a particular solution φ ( x ) for the equation y 00 + 4 y = (1 + x ) e 2 ix . Here we have α = 2 i . The characteristic equation is r 2 + 4 = 0 which has roots r = ± 2 i . Then since α is one of two distinct roots, Case 2 says we should guess a solution of the form φ ( x ) = ( A 0 x + A 1 x 2 ) e 2 ix . Computing the derivatives we get φ 0 ( x ) = [ A 0 + (2 A 1 + 2 iA 0 ) x + 2 iA 1 x 2 ] e 2 ix and also φ 00 ( x ) = [(2 A 1 + 4 iA 0 ) + (8 iA 1 - 4 A 0 ) x + - 4 A 1 x 2 ] e 2 ix .

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Plugging these into our differential equation and canceling the exponentials we get [(2 A 1 + 4 iA 0 ) + (8 iA 1 - 4 A 0 ) x - 4 A 1 x 2 ] + 4[ A 0 x + A 1 x 2 ] = 1 + x Collecting up like powers of x we get (2 A 1 + 4 iA 0 ) + (8 iA 1 ) x = 1 + x. Now we can equate the coefficients of corresponding powers of x to get (8 i ) A 1 = 1 coefficients of x (1) 2 A 1 + 4 iA 0 = 1 constant terms (2) Equation (1) gives us A 1 = 1 8 i = - i 8 and plugging this into equation (2) gives A 0 = 1 - 2 A 1 4 i = 1 16 - 1 4 i Thus we have φ ( x ) = 1 16 - 1 4 i x + - i 8 x 2 e 2 ix Since cos(2 x ) is the real part of e 2 ix our particular solution ψ will be the real part of φ . To find the real part, we must put φ in a + bi form: φ ( x ) = 1 16 - 1 4 i x + - i 8 x 2 e 2 ix = 1 16 - 1 4 i x + - i 8 x 2 (cos(2 x ) + i sin(2 x )) = 1 16 x cos(2 x ) + 1 4 x sin(2 x ) + 1 8 x 2 sin(2 x ) + i 1 16 x cos(2 x ) + 1 8 sin(2 x ) Now since ψ is the real part of φ we have ψ ( x ) = 1 16 x cos(2 x ) + 1 4 x sin(2 x ) + 1 8 x 2 sin(2 x ) .
One last useful fact The above three cases allow us to deal with polynomials, exponential functions, trig functions, and products of these things. It should be noted that since the operator L [ y ] = ay 00 + by 0 + cy is linear, we can also deal with sums.
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