00:01
Hi, here our given differential equation is y double dash plus 4y equals to f of t and y of 0 equals to 0, y dash of 0 is equal to 0 and further 0 less than or equal to t less than or equal to 1.
00:22
So here applying the laplace transformation on both the side we have l of y double dash plus l of y double dash plus l of y equals to l of ft and further on simplifying this we have s square y of s minus s y of 0 minus y dash of 0 plus y of s equals to integration over 0 to 1 e to the power minus st of 0 d t plus integration over 1 to infinity e to the power minus st t of 0 d t plus integration over 1 to infinity e to the power minus st of t d t.
01:03
So here on further simplification, we have s square y of s minus s y of 0 minus y dash of 0 plus y of s is equal to here substituting the limit and simplifying further it will be 0, integration over 1 to e to the power minus s t multiplied with t d t.
01:34
So here for this integration we need to use integration by parts method.
01:39
So applying integration by parts method for the limit 1 to infinity and simplifying this.
01:45
This equation can be reduced to s square the y of s minus s y -of -0 plus y of s equals to a to the power minus s upon s.
01:59
Plus a to the power minus s upon s square.
02:03
Now here further, using this given equation, we have to apply initial conditions.
02:16
So, substituting the value, we have s squared y of s minus 0 minus 0 plus y of s equals to e to the power minus s t upon s plus a to the power minus s upon s squared.
02:30
This is st so here on further simplification we have s square plus 1 y of s equals to e to the power minus s upon s plus e to the power minus s upon s square now further on simplifying this we got our value of y of s equals to e to the power minus s upon s multiplied with s square plus one again plus e to the power minus s upon s squared multiplied with s square plus one.
03:07
Now here we will reduce this in terms of h and g.
03:11
So y of s equals to e to the power s h of s and plus here we have a to the power minus s g of s so here for h of s we will use this partial traction method.
03:26
So here one upon s multiplied with s square plus one is equal to a upon s plus b upon s plus c upon s square plus 1.
03:37
So here, using partial fraction method and comparing, we have 1 equals to 8 times s squared plus 1 plus b of s plus c multiplied with s...