00:01
Hi, today we are solving the question in which we are given with the second order differential equation for x of t.
00:09
So d square x by dt square plus 5 into dx by dt plus 6x is equals to 3 where t is greater than 0 and given that x naught is equals to 0 and dx by dt naught is equals to 1 x dash naught is equals to 1.
00:37
So here taking laplace transform of both sides so l of d square x by dt square plus 5 into dx by dt plus 6x equals to 3.
00:53
Now here also we can write it as is equals to l of 3.
01:08
Now applying the laplace properties to the left side of the equation so we can write it as l of d square x by dt square plus 5 into l of dx by dt plus 6 into l of x is equals to 3s raised to the power 0.
01:34
Now from here applying initial condition and transform formulas so s square x of s minus x x naught minus x dash naught plus 5 into s of x x x of x s minus x naught plus 6 x s is equals to 3.
02:12
Now from here applying the initial conditions so s square x of s minus s naught minus 1 plus 5 into s of x s minus 0 plus 6 x s is equals to 3.
02:38
Simplifying the equation so s square x of s minus 1 plus 5 s x of s plus 6 x of s is equals to 3.
02:54
So here factoring out x s and solve it so we get x of s into s square plus 5 s plus 6 is equals to 1 plus 3.
03:09
So we can write x of s as equals to 4 by s square plus 5 x plus 6.
03:17
Now from here performing partial fraction decomposition x of s is equals to 4 by s plus 2 into s plus 3.
03:30
So it is can be written as a by s plus 2 plus b by s plus 3...