We want to find the solution
to the given initial value problem.
If we take a look at the given differential equation,
notice how it's a linear second order homogeneous
differential equation with constant coefficients,
meaning it fits this form here,
and therefore we can find the general solution
using the characteristic equation given here
where the values of a, b, and c are the
constant coefficients in the differential equations.
Then, once we find the solutions to the
characteristic equation, or the characteristic values,
we can determine which form
we'll use for the general solution,
and then from there we can find
the particular solution to this initial value problem.
So, again, looking at the given differential equation,
notice that a is equal to one, b is equal to four,
and c is equal to three, which means that
the characteristic equation would be r squared,
plus four r, plus three equals zero.
We can solve this quadratic by factoring.
It will factor into two binomial factors.
The factors of r squared are r and r.
The factors of positive three that add up to
positive four are positive three and positive one,
so this product is equal to zero
when r is equal to negative three,
or r is equal to negative one.
Notice how we have two distinct real roots,
which helps us determine which form
we use for the general solution.
For a quick review, if the
characteristic equation has two distinct real roots,
we use this form of the general solution.
If we have two real but equal roots,
we use this form for the general solution.
If we have complex roots, we use
this form for the general solution.
So in our case, for this example,
we're using the form from number one.
Let's go ahead and let r sub one be equal to negative three,
and r sub two be equal to negative one.
The general solution would be
y of x equals c sub one, a constant,
times e raised to the power of r sub one times x,
that would be negative three x,
plus c sub two times e raised to
the power of r sub two times x,
which would be negative one times x, or negative x.
That would solve the initial value problem.
We still have to use our initial conditions here,
y of zero equals two
and y prime of zero equals negative one,
to determine the values of c sub one and c sub two.
But to do this we'll also have to find the first derivative,
so let's go ahead and do that now.
Y prime of x would be equal to c sub one,
times e to the power of negative three x,
times negative three, or negative three c sub one,
e to the power of negative three x, plus c sub two,
times e to the power of negative x, times negative one,
or minus c sub two, e to the negative x.
Since we know that y after zero equals two,
we can use our function y, substitute zero for x,
and know that it's equal to positive two.
So, if x equals zero we would have c sub one,
times e to the power of negative three times zero,
that's zero, plus c sub two,
times e to the power of zero, must equal two.
This gives us the equation c sub one,
plus c sub two equals positive two,
again, using the initial condition and our function y.
Now, we also know that y prime of zero equals negative one,
so if y prime of zero equals negative one,
we can now use our derivative function
to determine a second equation involving
c sub one and c sub two.
So, if y prime of zero equals negative one,
we'll substitute zero for x again,
and now the derivative function
must be equal to negative one.
So, if x is zero, and e to the zero is equal to one,
we would have negative three times c sub one,
minus c sub two equals negative one.
Notice now we have two equations
that involve c sub one and c sub two,
so now we can solve for c sub one and c sub two
by solving this as a system of equations.
Let's go and do this on the next slide.
Again, we know c sub one plus c sub two
equals positive two and
negative three times c sub one,
minus e sub two equals negative one.
Notice how the c sub two terms are opposites,
so let's go ahead and solve this system using addition,
so we'll add these two equations together.
Notice how the c sub two terms simplify to zero,
giving us negative two times c sub one,
must equal positive one.
Dividing both sides by negative two
gives us c sub one equals negative 1/2.
Now to find c sub two, let's substitute
negative 1/2 for c sub one in this first equation.
Again, we now know that c sub one is equal to negative 1/2,
so we have negative 1/2, plus c sub two equals positive two.
Adding 1/2 to both sides would give us
c sub two equals two plus 1/2, or 5/2.
Now that we know the value of c sub one and c sub two,
we now have enough information to find
the solution to the given initial value problem.
Remember, we already found the general solution
was y of x equals c sub one,
times e to the power of negative three x,
plus c sub two, times e to the power of negative x,
and now we know the values of c sub one and c sub two,
so the solution to the initial value problem is
y of x equals negative 1/2,
times e to the power of negative three x,
plus 5/2, times e to the power of negative x.
Now, to help us verify that this solution is correct,
let's go ahead and graph this on the coordinate plane,
and because y of zero equals two,
this function must pass through the point zero, two.
Notice, here is the point zero, two.
Here's the graph of our solution,
and it does pass through the given point.
I hope you found this explanation helpful.