And we need to know another solution which is linearly

independent from this y1, okay?

The second solution we can obtain by the reduction of order,

say y2, which is of the form y1 times some another

function, g(x), which is not a constant.

Through the necessary computation in this case then,

we know that y sub 2 is equal to e to the r1x times g(x),

and the g(x) is given by the formula.

And because r1 is equal negative b over 2a,

plugging that onto this form and computing this intergration,

then we are going to get, this will be equal to x, right?

With the choice of suitable constant, so then it is what?

y2, which is x times e to the r1 of x,

which is linearly independent of y1 and

also satisfies the differential equation.

Because y1 and the y2 are linearly independent,

our general solution of differential equation will

be e to the arrow and x times c + 1 + c2 where c1 and

c2 are two of the equal constants, right?

Finally the case three, so

case three when the discriminate b case minus 4ac is negative.

We have two complex conjugates root sub categoristic equation.

So the two roots are r1 = alpha + i beta and

r2 = alpha minus i beta, where the alpha is real and

the beta is some positive constant, right?

Then y sub 1, which is e to the alpha + i beta x, and

the y sub 2, e to the alpha minus i beta times x.

They're two linearly independent but

complex-valued solutions of the differential equation.

But we prefer to have real-valued solutions,

because our original differential equation is a real coefficient,

real constant coefficient, second order homogenous differential equation.

Let me remind you here, the differential equation

one is ay double prime + by prime + cy = 0.

That's a differential equation.

Where the a is any non-zero real number, b and c are the real numbers, right.

So starting from this real constant coefficients,

second differential equation,

we prefer to have a real-valued solution.

But we get two linearly independent solution which is a complex valued.

So let's take a real and imaginary part of the solution y sub 1.

Say because this is a complex value of something it's a real part.

Real part of y sub 1 is equal to y1 + complex conjugate of y1 over 2.

That gives us the real part of it.

But in fact complex conjugator y1, that is equal to exactly y2.

You can confirm it by looking at this y sub 1 and the y sub 2.

Y sub 2 is a complex conjugate of y sub 1.

And what is it?

Using the Euler's formula, if you expanded this expression out,

then you can extract just the real part and the imaginary part.

Just the real part will be e to the alpha x times cosine beta of x.

By the similar token, this imaginary part of y sub 1,

which is y sub 1 minus complex conjugate y sub 1 over 2i,

which is the same as y sub 1 minus y sub 2 over 2i.

That is, in fact, e to the r of x times the sin beta of x, right?

Look at this expression in the middle, okay?

Linear combinations.

Real part of y1 and imaginary part of y1

are linear combination of two solutions y1 and

y2 linear combinations.

Then by the superposition principle for the homogeneous differential equation,

because both the y1 and the y2 are solutions of this differential equation.

Their linear combination, in fact which is a real part of y sub 1,

is also a solution of the same differential equation.

And the same is true for the imaginary parts.

In other words, e to the alpha x cosine beta of x.

And e to the alpha x times sine beta of x.

They're all solutions of this given second order

homogeneous differential equation, okay?

By the superposition principle.

Moreover, they're real valued, both functions

are real valued now and they are linearly independent.

You can check it simply by computing first kin of these two functions.

So we know have a what, two real valued linearly

independent solutions of the given differential equation.

And then this is general solution,

will be just a linear combination of this and that,

which we can write it as, because e to the alpha x is always common.

So y is equal to e to the alpha x times c1 cosine

beta of x + c2 sine beta of x, right?

This is the general solution of the given equation on n,

the b squared minus 4ac is strictly negative, right?