0:03

Yeah, we now handle the problem of finding a general solution for

Â homogeneous linear differential equation,

Â okay, with constant coefficients, okay?

Â With constant coefficients.

Â To make things a lot simple,

Â we restrict our service to the case of the order two.

Â So consider second order homogeneous linear

Â equation with constant coefficients which I write

Â it as Ay double prime + by prime + c is equal to 0.

Â Where the a is a non-zero constant and b and c they are all real constants.

Â So the problem we are concerned for the time being is the constant

Â coefficients second order homogeneous differential equation.

Â A y double prime + b y prime + c is equal to 0.

Â By inspection, we may guess that that differential

Â equation may have a solution of the form of e to the r of x.

Â What's the special property of this exponential function?

Â Its derivative is r times e to the rx.

Â Its second derivative will be r squared times e to the r(x), right?

Â And so on, okay?

Â So let us say y is equal to e to the rx in the differential equation 1.

Â Here is the differential equation one.

Â ay double prime + b prime + cy = 0,

Â a is a non-zero constant, and b,

Â c are arbitrary real constants, right?

Â In this equation, set y is equal to e to the rx, right?

Â Set y is equal to e to the rx.

Â 2:23

Is derivative is r times e to the rx,

Â second derivative will be r squared times e to the r of x.

Â So, in all those y or y prime or yw prime they always

Â contain the term e to the rx as a common factor.

Â So take this common factor out, e to the rx,

Â then the other parts will be ar squared + br + c.

Â And that must be equal to zero if this is the solution to that

Â differential equation.

Â That means what?

Â For each of the rx to be a solution of this second order,

Â constant of coefficient, differential equation,

Â r must be here, root of that quadratic equation, okay?

Â ar squared + br + c = 0, okay?

Â This quadratic equation we call it as the characteristic equation or

Â the auxiliary equation corresponding to the differential equation 1, okay?

Â So, in a sense solving this differential equation is reduced

Â to two algebraic problems, solving the root of this quadratic equation,

Â say the characteristic equation of the differential equation, okay?

Â As you know, depending on the choice of the constant a, b,

Â and c the type of roots of this quadratic equation will be different.

Â 4:20

Say ar squared + br + c = 0.

Â Try to solve it then.

Â We know that the root of this equation will be r

Â is equal to 1 over 2a times negative b plus or

Â minus square root of b squared minus 4ac.

Â And their nature depends on the sign of the so-called discriminant.

Â The quantity of this is square root the same, so for

Â discriminant of the quadratic equation, b squared minus 4ac and

Â we know where the characteristic equation 2.

Â This characteristic equation has two distinct real roots,

Â r1 and r2 when b squared minus 4ac is strictly positive.

Â It is one double real root,

Â say r1 is equal to r2 when the b squared minus 4ac = 0,

Â and it has a two complex conjugate roots say,

Â alpha plus minus beta, where alpha and beta are real constants.

Â In particular we can choose beta to be a positive constant, okay?

Â When b squared minus 4ac is a negative, right?

Â So we have three distinct cases for

Â the roots of the characteristic equation.

Â 5:59

So let's consider those three cases separately.

Â First, the e distinct case.

Â When we have two distinct real roots.

Â When this characteristic equation has two distinct real roots r1 not equal to r2.

Â Then e to the r1x is the solution, right?

Â And er2x is also a solution and

Â because r1 is not the same as r2,

Â you can compute distance functions,

Â and you can come from the Ruskin is never zero.

Â So then there is e to the r1 x and the e to the r2 of x,

Â they form also called fundamental set of solutions of differential equation.

Â And that means the general solution of

Â differential equation will be y = c1 times

Â e to the r1 of x and plus c2e times r2x.

Â That's the easiest case.

Â The second case, we have only one real double

Â root because b squared minus 4ac = 0.

Â So when this characteristic equation two has only one double root,

Â r1 = r2, then number must be through the roots formula, okay?

Â That number must be minus b/2a, right?

Â And we know then y1 = e to the r1 of x is a solution, right?

Â 7:46

And we need to know another solution which is linearly

Â independent from this y1, okay?

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

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

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

Â Through the necessary computation in this case then,

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

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

Â And because r1 is equal negative b over 2a,

Â plugging that onto this form and computing this intergration,

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

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

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

Â which is linearly independent of y1 and

Â also satisfies the differential equation.

Â Because y1 and the y2 are linearly independent,

Â our general solution of differential equation will

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

Â c2 are two of the equal constants, right?

Â Finally the case three, so

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

Â We have two complex conjugates root sub categoristic equation.

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

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

Â the beta is some positive constant, right?

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

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

Â They're two linearly independent but

Â complex-valued solutions of the differential equation.

Â But we prefer to have real-valued solutions,

Â because our original differential equation is a real coefficient,

Â real constant coefficient, second order homogenous differential equation.

Â Let me remind you here, the differential equation

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

Â That's a differential equation.

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

Â So starting from this real constant coefficients,

Â second differential equation,

Â we prefer to have a real-valued solution.

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

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

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

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

Â That gives us the real part of it.

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

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

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

Â And what is it?

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

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

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

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

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

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

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

Â Look at this expression in the middle, okay?

Â Linear combinations.

Â Real part of y1 and imaginary part of y1

Â are linear combination of two solutions y1 and

Â y2 linear combinations.

Â Then by the superposition principle for the homogeneous differential equation,

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

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

Â is also a solution of the same differential equation.

Â And the same is true for the imaginary parts.

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

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

Â They're all solutions of this given second order

Â homogeneous differential equation, okay?

Â By the superposition principle.

Â Moreover, they're real valued, both functions

Â are real valued now and they are linearly independent.

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

Â So we know have a what, two real valued linearly

Â independent solutions of the given differential equation.

Â And then this is general solution,

Â will be just a linear combination of this and that,

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

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

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

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

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

Â