We want to solve the differential equation
y to the sixth x dy/dx equals one plus x.
We want to use the initial condition y of one equals two
to express y to the seventh in terms of x.
The given differential equation
is a separable differential equation,
where all separable differential equations can be written
in the form of dy/dx equals
a function of x times a function of y.
But to solve the differential equation,
we want this equation to be in the form of
a function of y times dy
equals a function of x times dx.
Once we have it in this form,
we can integrate both sides of the equation
to find the general solution.
In this case, once we have the general solution,
we can find the particular solution
using the initial condition.
Starting with the given differential equation.
We want to have all the x terms on the left
and the y terms on the right.
So let's begin by dividing both sides by x.
Simplifying, we have y to the sixth dy/dx
equals a quantity one plus x divided by x.
Now, let's multiply both sides of the equation by dx,
so we'd have y to the sixth dy equals
quantity one plus x divided by x dx.
Notice now we do have a function of y times dy
equals a function of x times dx,
but before we integrate,
let's change the form of the right side of the equation.
Let's write this as y to the sixth dy equals
one over x plus x over x,
which we can rewrite or simplify
to just one over x plus one.
We have y to the sixth dy equals
the quantity one over x plus one dx.
Now, we'll integrate both sides of the equation.
We'd have the integral of y to the sixth dy
equals the integral of
one over x plus one dx.
Integrating the left side with respect to y,
we'd have y to the seventh divided by seven plus c.
We'll also have a constant of integration on the right,
so we'll only put plus c on the right.
We'd have y to the seventh divided by seven equals,
and now we integrate with respect to x.
The integral of one over x with respect to x is
natural log absolute value of x.
The integral of one with respect to x would be x,
so we have x, and let's write plus c sub one
for the constant of integration.
We actually don't need the absolute value here around the x,
because notice how we're told x is greater than zero.
To solve this for y to the seventh
and to find the general solution,
notice how we have to multiply
both sides of the equation by seven.
We'd have y to the seventh equals seven times
natural log x
plus x plus c sub one.
Distributing, we'd have y to the seventh equals
seven natural log x
plus seven x,
and then we'd have seven times c sub one,
which is just another constant.
Let's let c be equal to seven times c sub one,
so we'd have plus c.
Again, this would be the general solution,
but because we have an initial condition,
we can now find the particular solution.
Let's continue on the next slide.
The initial condition is y of one equals two,
so we can find the value of the constant
and the particular solution
by substituting one for x and two for y.
So we'd have two to the seventh
equals seven times natural log one
plus seven times one
plus c.
When natural log one is equal to zero,
so this term drops out.
To solve for c, we'd have two to the seventh
minus seven equals c.
Two to the seventh is equal to 128,
so we have 128 minus seven equals c,
so c is equal to 121.
By substituting 121 for c into the general solution,
we now have the particular solution.
We have y to the seventh equals seven
times natural log x
plus seven x
plus 121.
This is the particular solution
this question is asking for.
Again, this is the particular solution,
and this is the general solution.
I hope you found this helpful.