>> Welcome to the Cypress College Math Review
on Second Order Differential Equations, Special Types.
In general a 2nd order differential equation can be
written as a function of the independent variable,
the dependent variable and the first and second derivatives.
There are two special types of 2nd order differential equations
where a substitution reduces them
to a 1st order differential equation
that we're able to solve.
The first type has no dependent variable or no x in this case.
By substituting v equal dx dt or the first derivative,
the second derivative becomes the derivative of v
with respect to t or v prime.
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In this first example t is the independent variable
and x is the dependent variable and x is missing.
This is a 2nd order differential equation.
So we let v equal dx dt which means that the second derivative
of x with respect to t will equal dv dt or v prime.
We make these substitutions.
So t squared times v prime plus v squared is equal
to 2t times v. Interchanging the last two terms.
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Divide each term by t squared.
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The left hand side looks like we have a 1st order linear
differential equation.
But we look at the right hand side and we recognize that no,
we have instead a Bernoulli differential equation.
Here's the form for a Bernoulli differential equation.
So the first step to solve a Bernoulli differential equation
would be to divide by y to the nth
which in this case is v squared.
So we have v to the negative 2, v prime minus 2 over t,
v to the negative 1 is equal to negative 1 over t squared.
We make our substitution.
We let u equal y to the 1 minus n power.
Well in this case u will be v to the 1 minus 2 power.
So that's v to the negative 1 power.
We also have to find out what u prime is,
so u prime would be negative v
to the negative 2 times v prime by the chain rule.
So we replace v to the negative 2 v prime
with the opposite of u prime.
We replace v to the negative 1 with u.
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Multiply both sides of this equation by negative 1.
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Now we have a 1st order linear differential equation that's
in standard form.
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Now we solve this 1st order linear differential equation.
Here's our format.
We need to find the integrating factor mu.
There's our formula for mu.
P of x in this case is positive 2
over t. We simplify to get t squared.
So when we solve a 1st order linear differential equation we
get mu times y equals the antiderivative
of the product of q with mu.
So in this case we'll have the integrating factor mu
which is t squared times our function u is equal
to the antiderivative of mu
which is t squared times the function q which is 1
over t squared, we're integrating that with respect
to t. So we get t plus the constant of integration.
So u is equal to t plus c sub 1 over t squared.
And now we substitute back, u is equal it v the negative 1 power
so 1 over v is equal to this.
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So let's take reciprocals.
So v is equal to t squared over t plus c sub 1
and we substitute back.
Recall that v is equal to dx dt.
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We're going to need to integrate that.
So since the degree on top is greater than or equal
to the degree on the bottom, we need to do long division.
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So t squared divided by t is t. So t times t is t squared,
c sub 1 times t, change the signs, divide negative c sub 1 t
by t and I get negative c sub 1.
Multiply negative c sub 1 times 9 divisor
and I get negative c sub 1t minus c sub 1 squared.
Change the signs and I get c sub 1 squared.
So I'm integrating 1dx equals and when I'm integrating
over here on the right side is my quotient
which is t minus c sub 1 plus my remainder
which is c sub 1 squared over my divisor which is t plus c sub 1.
I'm integrating with respect to t.
So on the left hand side I get x,
on the right hand side I get t squared
over 2 minus c sub 1 times t plus c sub 1 squared times the
natural log of the absent value of my denominator
which t plus c sub 1 plus a new constant of integration.
And notice that x equals another constant is also a solution.
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Here's another 2nd order differential equation.
The independent variable is t,
the dependent variable is x and x is missing.
When you look at this term here that's not the 2nd derivative,
that's the second power of the 1st derivative.
So we make our substitutions.
V is equal to dx dt and the 2nd derivative is dv dt or v prime.
So we get 2t times v prime equals v squared minus 1.
This is separable differential equation.
So we have 1 over the quantity v squared minus 1 dv equals 1
over the quantity 2t dt.
To be able to integrate that we have do a partial
fraction decomposition.
So we have 1 over the product of v plus 1 with v minus 1.
To set that up we would have a over b plus 1 plus b
over v minus 1, multiply both sides
by the least common denominator
which is the original denominator.
So we get 1 equals a times v minus 1 plus b times v plus 1,
let v equal 1 and we get that 1 is equal to 2 times b
which gives that b is equal to 1/2.
If we then let v equal negative 1 we get that 1 is equal
to negative 2a which tells us that a is equal to negative 1/2.
So now the integral on the left hand side becomes a
which is negative 1/2 over v plus 1 plus b
which is 1/2 over v minus 1.
On the right hand side we're integrating 1/2 over t dt,
multiply by 2 and we get negative natural log
of the absolute value of the denominator plus natural log
of the absolute value of the denominator equals natural log
of the absolute value
of the denominator plus our constant of integration.
We're going to go ahead and call our constant
of integration the natural log of the absolute value of c
so that we're able to combine these logarithms.
That's a trick that you can use
that makes things easier for you.
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Let's combine the logarithms.
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Exponentiate both sides.
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So now we have this quotient equals plus
or minus c t. We can go ahead and call
that a new constant c sub 1 times t. Now let's solve for v.
So we cross multiply and we have v minus 1 equals c sub 1 tv plus
c sub 1 t. So v minus c sub 1 tv equals c sub 1 t plus 1.
Factor out a v.
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So v is equal to c sub 1 t plus 1 divided
by 1 minus c sub 1 times t. Let's substitute back.
V is equal to dx dt, separate.
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And since the degree on top is greater than or equal
to the degree on the bottom we have to do long division.
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Divide c sub 1t by negative c sub 1t
and we simply get negative 1.
Multiply negative 1 times my divisor
and we get c sub 1 times t minus 1, change the signs
and we get 2 as our remainder.
So on the right hand side we're integrating negative 1
which is our quotient plus or remainder
which is 2 over our divisor.
So our answer x is equal to negative t plus and we need
to do a u substitution here to be able to integrate this.
U equals negative c sub 1t plus 1, so du is negative c sub 1 dt.
So our integral turns out to be negative 2
over c sub 1 times natural log of the absolute value
of 1 minus c sub 1 times t plus the constant of integration.
That's the answer to our differential equation.
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Our second type of 2nd order differential equation,
a special type has no independent variable,
t is missing.
In this case we replace dx dt with v and our goal here is
to make t completely disappear.
So we have x in our equation.
We're replacing dx dt with v. Now we need
to replace the 2nd derivative with something
that has no t's in it at all.
So the 2nd derivative is equal to dv dt but we need that t
to be completely gone.
We can't just call it v prime
because v prime is the same as dv dt.
We need that t to be out of the equation completely.
By the chain rule dv dt is equal dv dx times dx dt.
Dx dt recall is equal to v. So we make that substitution.
So the 2nd derivative can be written as dv dx times v.
If we make that substitution now our equation can be written
in terms of x v and dv dx times v. No t's at all.
Now we can solve the differential equation.
Here's our first example.
It's a 2nd order differential equation and t is missing.
So we make our substitutions.
We're going to replace dx dt with v and then we're going
to replace the 2nd derivative with dv dx times v.
So x squared plus 1 times dv dx times v equals 2x times
v squared.
This is a separable differential equation.
I get 1 over v dv equals 2x
over the quantity x squared plus 1 dx.
So natural log of the absolute value of v equals natural log
of the absolute value of the quantity x squared plus 1 plus
and then I chose the form of my constant of integration
to be natural log of the absolute value of c
because all the other terms are natural logs.
So this is equal to natural log of the absolute value
of c times the quantity x squared plus 1.
I combine those logarithms on the right hand side,
exponentiate both sides.
So I have absolute value of v equals absolute value
of c times the quantity x squared plus 1.
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So v is equal to plus or minus c times the quantity x squared
plus 1.
So v is equal to a new constant times that quantity.
Replace v to what it's equal to which is dx dt.
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And is a separable differential equation so we have 1
over the quantity x squared plus 1 dx.
And the right hand side we're just integrating the constant.
So on the left hand side we get arctangent,
the right hand side we get c sub 1 times t plus a new constant
of integration.
So x equals tangent of this quantity.
X equals a constant is also a solution
to this differential equation.
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And that gives us the solutions to this differential equation.
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On this next example we don't even see what our independent
variable would be.
Let's write it in terms of [inaudible] notation
and just choose an independent variable.
So it would be the 2nd derivative of y,
let's say with respect to t plus y times the 1st derivative cubed
equals zero.
So the independent variable t is missing,
we're going to make our substitution.
So v is equal to dy dt and the 2nd derivative is equal
to dv dy times v. So we replace the 2nd derivative
with v times dv dy.
And then we have plus y times and the derivative gets replaced
with v, so we have the derivative cubed
so we have v cubed and that's equal to zero.
This is a separable differential equation separating the terms.
We have v over v cubed dv equals negative y dy.
On the left hand side we're integrating v to the negative 2.
On the right hand side we're integrating negative y.
So we get negative v
to the negative 1 equals negative y squared
over 2 plus c sub 1.
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So we get 1 over v is equal to, to multiply by negative 1
but let's also get a common denominator here.
So 1 over v is equal
to y squared plus new constant over 2.
So v is equal to 2 over y squared plus c sub 2.
Now we substitute back.
So v is equal to dy dt.
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So we integrate the left hand side,
we're integrating y squared plus c sub 2 with respect to y.
And the right hand side we're integrating just 2.
So we have y cubed over 3 plus c sub 2 times y equals 2t plus
another constant c sub 3 or a constant of integration.
If I multiply 3 by 3 we have y cubed plus 3 times cu sub 2,
so we have another constant, a new constant,
we'll call it c sub 4 times y equals 6t plus 3 times our old
constant c sub 3, we'll call that c sub 5.
So here's our solution to the differential equation.
Now you do need to note that you have two different arbitrary
constants here.
I call them c sub 4 and c sub 5.
Your numbering scheme will probably be different.
But you do have to note that you do have two different arbitrary
constants here.
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Here's some additional problems to try on your own.
I encourage you to pause the video, do these problems
and then check with the solutions that are
at the end of the video.
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Here are the solutions to the practice problems.
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