00:01
In this problem we have equation, so we have to solve this 3x minus 1 by e raised to the power to y, dy plus dx is equals to 0.
00:18
Okay.
00:19
So first of all we will substitute y is equals to e raised to the power to y.
00:26
Okay.
00:28
Substituting u is equals to e raised to the power to y so we will differentiate this d u is equal to e raised to the power to y so if we differentiate this so two will come forward two e raised to the power two y d y so d y is equals to d u divided by two e raised to the power 2y.
01:02
This is our first equation.
01:05
Okay.
01:05
Now dx plus we have dx here plus 3x minus 1 by 4.
01:23
The whole term d u divided by 2 u.
01:32
This is equals to 0 because we have substitute u is equal to e raise to the power 2.
01:39
So let's solve this 3x by 2 u minus 1 by 2 u square into d u is equals to d x will be minus d x in right direction okay now we will reduce this to homogeneous substitution method okay so homogeneous substitution which is u is equals to z lambda z raised to the power lambda so lambda into z raised to the power lambda minus 1 into will take the whole term 3 into z raised to the power 1 minus lambda divided by 2 minus 1 by 2 into z raised to the power 2 lambda into d z which is equals to minus d z okay since we have taken x is equals to z u is equals to z less to the power lambda so value is 0 is equals to minus lambda minus 1 is equals to 0.
03:15
So lambda is equals to minus 1.
03:18
This is the value of lambda.
03:23
Now we will substitute this...