00:01
In this problem we are provided with the differential equation t times d u over d t equals to t squared plus three times u.
00:15
Here t is greater than 0 and we are provided with the initial condition that u of 5 equals to 100.
00:25
So let us rearrange the differential equation a little by dividing with t first we get d u over d t to be equal to t plus 3 times u over t.
00:36
Now subtracting with 3 times u over t, we have du over dt minus 3 times u over t equals to t.
00:44
Comparing this with the bernouri's form, that is du over dt plus p times u equals to q, we obtain the value of p to be negative 3 over t and the value of q to be t.
01:02
And the equation of the bernanis differential equation is given by u times the integrating factor which equals to the integral of q times the integrating factor d t where the integrating factor is equal to e raised to the power of integral of p d t.
01:27
So let us begin by finding out the integrating factor.
01:30
So we have integrating factor to be equal to e raised to the power of integral of negative 3 over t dt.
01:39
Let us evaluate this.
01:42
We get e raised to the power of negative 3 times the integral of 1 over t which is natural log of t.
01:50
This can be rewritten as e raised to the power of natural log of t raised to the power negative 3 by the power rule of natural logarithm.
01:59
Since e and natural log are inverses, we get t raised to the power negative 3, which is equal to 1 over t cube...