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diffeequatios2

# diffeequatios2 - 1 Solving Differential Equations by the...

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Unformatted text preview: 1 Solving Differential Equations by the Laplace Transform T¨urkay Yolcu 1 22nd March 2012 Details are left to the reader. Please follow the steps to figure out the complete solution. Note that differentiating L{ x ( t ) } ( s ) = Z ∞ e- st x ( t ) dt with respect to s we get L{ tx } =- d ds L{ x } , L{ t 2 x } =- d ds L{ tx } =- d ds- d ds L{ x } = d 2 ds 2 L{ x } An immediate upshot of this formula is the following: L{ t cos( kt ) } =- d ds s s 2 + k 2 = s 2- k 2 ( s 2 + k 2 ) 2 Moreover, using the integration by parts we obtain L{ x } = s L{ x } - x (0), L{ x 00 } = s 2 L{ x } - sx (0)- x (0). For example, observing that 1 ( s 2 + 9) 2 = 1 18 18 ( s 2 + 3 2 ) 2 = 1 18 ( s 2 + 3 2 )- ( s 2- 3 2 ) ( s 2 + 3 2 ) 2 = 1 18 1 s 2 + 3 2- ( s 2- 3 2 ) ( s 2 + 3 2 ) 2 gives L- 1 1 ( s 2 + 9) 2 = 1 54 sin(3 t )- 1 18 t cos(3 t ) Problem 1. Solve for x ( t ) and y ( t ), given that dx dt + x + 4 y = 10, x- dy dt- y = 0, x (0) = 4, y (0) = 3 Solution: Taking the Laplace transform of each equation yields: dx dt + x + 4 y = 10 ⇒ sX- x (0) + X + 4 Y = 10 s ⇒ ( s + 1) X + 4 Y = 10 s + 4 x- dy dt- y = 0 ⇒ X- ( sY- y (0))- Y = 0 ⇒ X- ( s + 1) Y =- 3 ⇒ ( s + 1) X- ( s + 1) 2 Y =- 3( s + 1) 1 T. Yolcu, Department of Mathematics at Purdue University. Email:T....
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diffeequatios2 - 1 Solving Differential Equations by the...

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