- #1

Jag1972

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- TL;DR Summary
- Compare the solution of delayed forcing function differential equation using Laplace transform and method of undermined coefficients.

Tried to figure out myself but have now admitted defeat, requesting some guidance from you good people. Not looking for any specific answers, unless the problem is my working out and not my process.

If we take the following differential equation: ##y(t)'' + 4y(t) = 7u(t-2)## and determine solution for following time range: ##\:\:0\leq t \lt \infty \: ## with initial conditions: ##y(0) = 2, \:y'(0) = 4##.

First solved using Laplace Transforms (This is the most straight forward way), this is solution calculated.

##y(t) = u(t-2)(\frac {7}{4} - \frac {7}{4} cos (2(t-2))) + 2 cos (2t) + 2 sin (2t)##

I tried to get a solution that would match this using method of undermined coefficients. This was my approach which felt intuitive to me. Determined 2 separate solutions for the following differential equations:1) ##y(t)'' + 4y(t) = 0\:\:## for range ##\:\:0\leq t \lt 2 \: ## initial conditions ##y(0) = 2, \:y'(0) = 4##.

2) ##y(t)'' + 4y(t) = 7## for range ##\:\:0\leq t \lt \infty \: ## initial conditions ##y(2) = -2.8, \:y'(2) = 0.4##.

The 2 solutions obtained were:

1) ##y(t)=2 cos(2t) + 2 sin (2t)\:\:## for range ##\:\:0\leq t \lt 2 \: ##

The initial conditions for the second solution were determine using solution 1 and its derivative. Simply replaced t with 2 in both equations.

2) ##y(t)=(2.9) cos(2t) + (3.5) sin (2t) + \frac {7}{4}\:\:## for range ##\:\:0\leq t \lt 2 \: ##

My problem is that the Laplace transform and Undermined coefficients solutions don't give same results for ##t\geq2##.

for example when ##t=3##

1) Laplace Transform

##y(t) = u(t-2)(\frac {7}{4} - \frac {7}{4} cos (2(t-2))) + 2 cos (2t) + 2 sin (2t)##

## = u(1)(\frac {7}{4} - \frac {7}{4} cos (2(1))) + 2 cos (5) + 2 sin (5)##

2) Undermined Coefficients

##y(t)=(2.9) cos(2t) + (3.5) sin (2t) + \frac {7}{4}\:\:## for range ##\:\:0\leq t \lt 2 \: ##

##-0.7836=(2.9) cos(5) + (3.5) sin (5) + \frac {7}{4}\:\:## for range ##\:\:0\leq t \lt 2 \: ##

Would appreciate any support on this.

If we take the following differential equation: ##y(t)'' + 4y(t) = 7u(t-2)## and determine solution for following time range: ##\:\:0\leq t \lt \infty \: ## with initial conditions: ##y(0) = 2, \:y'(0) = 4##.

First solved using Laplace Transforms (This is the most straight forward way), this is solution calculated.

##y(t) = u(t-2)(\frac {7}{4} - \frac {7}{4} cos (2(t-2))) + 2 cos (2t) + 2 sin (2t)##

I tried to get a solution that would match this using method of undermined coefficients. This was my approach which felt intuitive to me. Determined 2 separate solutions for the following differential equations:1) ##y(t)'' + 4y(t) = 0\:\:## for range ##\:\:0\leq t \lt 2 \: ## initial conditions ##y(0) = 2, \:y'(0) = 4##.

2) ##y(t)'' + 4y(t) = 7## for range ##\:\:0\leq t \lt \infty \: ## initial conditions ##y(2) = -2.8, \:y'(2) = 0.4##.

The 2 solutions obtained were:

1) ##y(t)=2 cos(2t) + 2 sin (2t)\:\:## for range ##\:\:0\leq t \lt 2 \: ##

The initial conditions for the second solution were determine using solution 1 and its derivative. Simply replaced t with 2 in both equations.

2) ##y(t)=(2.9) cos(2t) + (3.5) sin (2t) + \frac {7}{4}\:\:## for range ##\:\:0\leq t \lt 2 \: ##

My problem is that the Laplace transform and Undermined coefficients solutions don't give same results for ##t\geq2##.

for example when ##t=3##

1) Laplace Transform

##y(t) = u(t-2)(\frac {7}{4} - \frac {7}{4} cos (2(t-2))) + 2 cos (2t) + 2 sin (2t)##

## = u(1)(\frac {7}{4} - \frac {7}{4} cos (2(1))) + 2 cos (5) + 2 sin (5)##

2) Undermined Coefficients

##y(t)=(2.9) cos(2t) + (3.5) sin (2t) + \frac {7}{4}\:\:## for range ##\:\:0\leq t \lt 2 \: ##

##-0.7836=(2.9) cos(5) + (3.5) sin (5) + \frac {7}{4}\:\:## for range ##\:\:0\leq t \lt 2 \: ##

Would appreciate any support on this.

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