00:01
And the question we are given that u on differentiating two terms with respect to x is equal to one divided by 16 u t t.
00:11
That's it is our first equation.
00:13
And we know that u of zero comma t is equal to u of y comma t, which is equal to zero and u of x comma zero is equal to 12 sin pi x minus 6 sin of 8 pi x and u t of x comma zero is equal to zero.
00:40
This method of separation of variables is to take the solution as form of product.
00:45
So let u of x comma t is equal to a of x multiplied with b of t.
00:55
And that is a solution of particular differential equation and putting it in equation first, we get del square divided by del x square, where del is of a at a of x multiplied with b of t, which is equal to one divided by 16 del square multiplied with del square divided by del t square where del is of again a of x multiplied with b of t.
01:31
And now for a double dash, we have a double dash of x multiplied with b of t is equal to one divided by 16 multiplied with a of x multiplied with b double dash of t.
01:51
And for average, we can write a double dash of x divided by a of x is equal to b double dash of t divided by b of t multiplying with one divided by 16 from which we can write one divided by 16 multiplied with what here, double dash of x divided by a of x is equal to b double dash of t divided by b of is equal to minus lambda square.
02:33
Let lambda square is in separation constant.
02:37
Now, since 16 multiplied with a double dash of x divided by a of x is equal to minus lambda square and b double dash of t divided by b of t is equal to minus lambda square, then 16 multiplied with a double dash of x is equal to minus of lambda square a x from which we can write 16 of 16 multiplied with a double dash of x is equal to lambda square minus lambda square multiplied with a of x and we can write this as 16 multiplied with a double dash of x plus lambda square a of x is equal to zero.
03:45
And for which the auxiliary equation auxiliary equation would be 16 m square plus lambda square is equal to zero for which we get m is equal to plus minus i of lambda divided by now our general solution which means a x is equal to c1 cos of lambda divided by 4 x plus c2 sin of lambda divided by 4 multiplied with x and then we are given that u of zero comma t is equal to zero and which means a of zero multiplied with b of t is equal to zero.
04:50
So, a of zero is equal to zero or b of t is equal to zero.
04:57
We will consider only a of zero is equal to zero since b of t is equal to zero then u then u of x comma t is equal to zero which is a trivial solution but we are searching for the non -trivial solution...