We're going to solve the separable differential
equation and also use the initial condition,
y of zero equals three to write the answer as a formula
in the variable t.
In order to solve a separable differential equation
we're going to manipulate the given equation
so it's in the form of a function of y times dy equals,
in this case, a function of t times dt.
Looking at the given differential equation,
let's begin by factoring denominator.
So we're going to have dy dt
equals t divided by,
again the greatest common factor of these two terms is y.
So we're going to have y times the quantity t squared plus one.
So if we want to have all the y terms on the left
and the t terms on the right,
we don't want this y here in the denominator,
so lets multiply both sides of the equation by y.
So we'd multiply the right side by y, the left side by y.
So we'd have y times dy dt equals t
divided by the quantity t squared plus one.
And now multiply both sides of the equation by dt
so we have y dy equals t
divided by the quantity t squared plus one dt.
Notice now we do have the equation in the form
where we have a function of y times dy
equals a function of t times dt.
So for the next step we integrate both sides of the equation
So we'd have the integral of y dy equals the integral of t
divided by the quantity t squared plus one dt.
Now integrating, on the left side we'd have
y squared divided by two plus c
Let's write that as one half y squared
because we'll also have a constant of integration
on the right, let's only put the constant on the right.
So we'd have one half y squared equals
and integrate here with the perform U-Substitution
or let the denominator be equal to u so we'd have u
equals t squared plus one, so differential u is equal to
two t dt and if we only have a t dt,
so if we divide both sides by two notice how one half
differential u is equal to t dt.
Which means we can think of all of this as just
one half times one over u du.
So the integral in terms of u would be one half times
natural log absolute value of u which means in terms of t
we'd have one half times natural log absolute value
of t squared plus one plus the constant of integration,
let's call it plus c sub one.
And notice how t squared plus one is always going to be
positive so we really don't need the absolute value here.
Now we're going to work on solving this for y
so for the next step let's multiply both sides by two
should give us y squared equals two times the quantity
one half natural log of the quantity t squared plus one
plus c sub one.
So distributing we have y squared equals natural log
of the quantity t squared plus one
and we really have plus two times c sub one
but it's just another constant.
Let's let c be equal to two times c sub one
so we just have plus c.
Let's continue solving for y in the next slide.
So now to solve for y we'd hit the square root of both sides
of the equation, we would normally have a plus or minus here
but we were told that y is greater than zero
so we can say that y equals the square root of natural log
of the quantity t squared plus one plus c, so this would be
the general solution to the given differential equation.
Now remember we want to find the particular solution
given the initial condition y of zero equals three.
So using the initial condition y of zero equals three
we can substitute zero for t, three for y and solve for c
which will give us the particular solution.
So making the substitution we would have three
equals the square root of natural log of t squared plus
what would be zero squared plus one plus c.
So we have three equals, this is going to be natural log one
which equals zero so we just have three
equals the square root of c, so to solve for c we would
square both sides of the equation which gives us c
equals nine.
So by substituting nine for c in the general solution
we now have the particular solution
given the initial condition.
So we would have y equals the square root of natural log
of the quantity t squared plus one plus nine.
Then again this is called the particular solution.
I hope you found this helpful.