We want to solve initial value problem
with the given separable differential equation.
We have a differential equation
y prime equals four x y minus two x,
if initial condition y of zero equals eleven.
All separable differential equation
can be written in the form of
dy dx equals a function of x times a function of y.
And then we can change the form of the equation
so we have a function of y times dy
equals a function of x times dx.
Once we have the equation in this form
we can integrate both sides of the equation
to find the general solution
and then once we find the general solution
we can find the particular solution using initial condition.
Let's rewrite the given differential equation.
Let's use dy dx instead of y prime
so the equation would be dy dx
equals
let's also factor the right side
the equation come in factor of two x.
So we'd have two x times the quantity two y minus one.
Notice in this form,
we do have dy dx equals a function of x
times the function of y.
Now let's write this in differential form
or multiply both sides by dx.
So we'd have differential y equals two x
times the quantity two y minus one
times dx.
Remember our goals to have a function of y times dy
equals the function of x times dx.
So want the quantity two y minus one
to be on the left side of the equation.
So let's divide both sides with the equation
by the quantity two y minus one.
So let's rewrite this as
one
over the quantity two y minus one differential y
equals
this simplifies out we'd just have
two x dx on the right.
So again now we have a function of y times dy
equals a function of x times dx.
So now integrate both sides of the equation.
So we have the integral
of one over the quantity two y minus one dy
equals the integral
two x dx.
We need to perform u substitution on the left side
where u
is equal to
two y minus one.
So differential u is equal to two dx.
So if we divide both sides by two,
notice it one half du equals dx.
Which means,
which means you can think of all of this
as
one half times one over u du.
So the integral with respect to u would be
one half natural log absolute u.
Which means in terms of y,
we'd have one half times natural log absolute value
of the quantity two y minus one
and then plus the constant of integration
but we'll also have a constant on the right,
so let's only put the constant on the right.
So on the right side,
integrating this pack to x would have
two times x squared divided by two plus c
or just x squared
let's call it plus c sub one.
Now we need to solve this for y.
Let's multiply both sides of the equation by two.
Which would give us natural log
absolute value of two y minus one
equals
two x squared plus two times c sub one.
Let's write that product as two x squared
plus c sub two
where c sub two equals two times c sub one.
Now let's write log equation as an exponential equation
where natural log is base e.
So the exponential equation we'd have e
raised to the power of two x squared plus c sub two
equals two y minus one.
[mumbles] two y minus one
equals
e raised to the power of two x squared plus c sub two.
Notice here we're adding exponents
so let's write this as a product.
We would have two y minus one equals
e raised to the power of two x squared times
e raised to the power of c sub two
but e raised to the power of c sub two
is just some other constant
so let's let c be equal to
e to the power of c sub two.
So we can write the right side
as just c times e raised to the power of two x squared.
And now we can solve this for y
and find the general solution.
So next happen to be add one to both side
so we'd have two y equals
c times e raised to the power of two x squared plus one.
And for last step we'll divide both sides by two.
So we have y equals c times e raised to the power
of two x squared plus one
all divided by two.
Again this is the general solution
to the differential equation.
Remember we're also given that y of zero
equals eleven.
So using this,
we can find the particular solution
by determining the value of c.
We'll substitute zero for x
and eleven for y.
So that'll give us the equation eleven
equals
c times e raised to the power of two times zero squared
plus one
all over two.
Let's clear the fraction here
by multiplying both sides by two.
Which we'll give us 22 equals
e to the zero is one
so on the right side we just have c plus one
subtracting one on both sides gives us c equals 21.
Now that we have the value of c,
we now know the particular solution.
The particular solution would be y equals
21
times e raised to the power of two x squared plus one
all divided by two.
We could also write this as
y equals 21 halves
e raised to the power of two x squared plus one half.
Either of this form are acceptable
for the particular solution.
I hope you found this helpful.