00:01
In this question, we are asked to solve the given equation.
00:03
And basically it's a differential equation with initial condition, with an initial condition.
00:09
We can rewrite the equation as 2t squared minus 2t xx2t times d theta over d t equals 5.
00:17
And then we can rewrite it as negative 2 t d theta over d t equals to 5 minus 2 t squared.
00:29
And after dividing both sides by negative 2t, we are going to get d theta over dt equals to 5 minus 2t squared divided by negative 2t.
00:47
Next, let if we integrate both side of the equation, we are going to get that theta equals to the integral of 5 minus 2t squared divided by negative 2t.
01:03
We can rewrite that at the sum of two integrals, the integral of 5 over negative 2 t, plus the integral of t, because negative 2t squared over negative 2 t equals to t.
01:28
In the first case, in the first integral equals to negative 5 over 2 times l .n of the absolute value of t.
01:35
The second integral is t squared over 2, and there should be a constant of integration c because the integrals are indefinitely...