00:01
In this question, we're going to solve the differential equation dy dx equals the quantity of 2y plus 3 over 4x plus 5 being squared.
00:10
So we're going to have to separate variables.
00:13
My first step is i am going to say if i have a fraction being squared, i get that by squaring each of the numerator and denominator.
00:24
So i can rewrite this right -hand side as 2y plus 3 quantity squared over 4x plus 5 quantity squared.
00:36
Now, i'm going to cross multiply.
00:39
When i do, i get 4x plus 5 quantity squared times dy is equal to the quantity of 2y plus 3 being squared times dx.
00:57
What do we do? we're going to divide the 2y plus 3 quantity squared to the left and divide the 4x plus 5 quantity squared to the right.
01:07
So we have 1 over 2y plus 3 quantity squared dy equals 1 over the quantity of 4x plus 5 being squared dx.
01:26
Equivalently, we have 2y plus 3 quantity to the negative second power dy equals 4x plus 5 to the negative second power, that whole quantity times dx.
01:44
Now we'll integrate each side.
01:46
My integral on the left, 2y plus 3 quantity to the negative first divided by negative one, divide all of that by two, by the coefficient on y on the right -hand side.
02:06
My antiderivative is the quantity of 4x plus 5 raised to the negative first over negative one.
02:16
Take all of that and divide it by four, the coefficient on x.
02:21
And then of course on that side, i need a plus c.
02:27
Now if i clean this up, what do we have? we're going to have negative one over twice the quantity of 2y plus 3 equals negative one over four times the quantity of 4x plus 5 plus c...