How to solve 3rd order Diff Eqn

[Lost souls]

Can someone tell me how to solve third order differential eqn and what is the general solution?? For example
(d³y/dx³) - 2(d^​2y/dx²) + 4(dy/dx) - 8y=0

 

[HallsofIvy]

That's a "homogeneous linear equation with constant coefficients" and the method of solution is pretty "cut and dried". It's "characteristic equation" is the cubic polynomial equation, \(\displaystyle r^3- 2r^2+ 4r- 8= 0\). That's fairly easy to solve. Notice that the coefficients are powers of 2. That should make you think to try r= 2 as a solution: \(\displaystyle 2^3- 2(2^2)+ 4(2)- 8= 8- 8+ 8- 8= 0\). So r= 2 is a solution and then, dividing by r- 2, we have \(\displaystyle r^2+ 4= 0\) which is the same as \(\displaystyle r^2= -4\) so that r= 2i and r= -2i are the other two roots of the characteristic equation. Do you know how to "assemble" the general solution to the differential equation from those?

 

[Lost souls]

That's a "homogeneous linear equation with constant coefficients" and the method of solution is pretty "cut and dried". It's "characteristic equation" is the cubic polynomial equation, \(\displaystyle r^3- 2r^2+ 4r- 8= 0\). That's fairly easy to solve. Notice that the coefficients are powers of 2. That should make you think to try r= 2 as a solution: \(\displaystyle 2^3- 2(2^2)+ 4(2)- 8= 8- 8+ 8- 8= 0\). So r= 2 is a solution and then, dividing by r- 2, we have \(\displaystyle r^2+ 4= 0\) which is the same as \(\displaystyle r^2= -4\) so that r= 2i and r= -2i are the other two roots of the characteristic equation. Do you know how to "assemble" the general solution to the differential equation from those?

the 'assembling' part is the problem, if that is what it's called. .in case of a DE of second order, for real and distinct roots, y = C₁ e^m₁x + C₂ e^m₂x
and for imaginary roots, y= e^ax [C₃ cos bx + C₄ ​sin bx]. .
also, as u said, by operator equivalent method , the equation is
D³ -2D²+ 4D -8 = 0;
(D² + 4)(D - 2) = 0;
D = +-2i or D = 2;

how to proceed forward. .the roots are neither real and distinct,nor real and same or completely imaginary. .thats the problem.

 

[Deleted member 4993]

the 'assembling' part is the problem, if that is what it's called. .in case of a DE of second order, for real and distinct roots, y = C₁ e^m₁x + C₂ e^m₂x
and for imaginary roots, y= e^ax [C₃ cos bx + C₄ cos bx]. .
also, as u said, by operator equivalent method , the equation is
D³ -2D²+ 4D -8 = 0;
(D² + 4)(D - 2) = 0;
D = +2i or D = -2i or D = 2;

how to proceed forward. .the roots are neither real and distinct,nor real and same or completely imaginary. .thats the problem.

Exactly like the 2nd Order case!

The roots are distinct (r1, r2 & r3 or 2, 2i and -2i). So the homogeneous solution is:

y_H = Ae^(r1*x) + Be^(r2*x) + Ce^(r3*x)

 

[HallsofIvy]

the 'assembling' part is the problem, if that is what it's called. .in case of a DE of second order, for real and distinct roots, y = C₁ e^m₁x + C₂ e^m₂x
and for imaginary roots, y= e^ax [C₃ cos bx + C₄ cos bx].

Well, that would be for the complex roots, a+ bi and a- bi. If they were pure imaginary, a= 0, you would have just \(\displaystyle C_3cos(bx)+ C_4sin(bx)\) because, with a= 0, \(\displaystyle e^{ax}= e^{0x}= e^0= 1\)

also, as u said, by operator equivalent method , the equation is
D³ -2D²+ 4D -8 = 0;
(D² + 4)(D - 2) = 0;
D = +-2i or D = 2;

how to proceed forward. .the roots are neither real and distinct,nor real and same or completely imaginary. .thats the problem.

??? You have found that the roots are 2, 2i, and -2i. That is precisely one real root and two conjugate imaginary roots. From what you say here, you should have \(\displaystyle y= Ae^{2x}+ Bcos(2x)+ Csin(2x)\)

 

[Lost souls]

Well, that would be for the complex roots, a+ bi and a- bi. If they were pure imaginary, a= 0, you would have just \(\displaystyle C_3cos(bx)+ C_4sin(bx)\) because, with a= 0, \(\displaystyle e^{ax}= e^{0x}= e^0= 1\)


??? You have found that the roots are 2, 2i, and -2i. That is precisely one real root and two conjugate imaginary roots. From what you say here, you should have \(\displaystyle y= Ae^{2x}+ Bcos(2x)+ Csin(2x)\)

i have learnt DE's upto second order only. .bt most of the questions in the exercise at the end of the topic are of third order. .i got used to having the two roots being either completely real or complex(or imaginary in certain cases) and then applying the general form of the solution. But 3 mixed roots, both real and imaginary, have me all at sea. .:-|

 

[Lost souls]

Exactly like the 2nd Order case!

The roots are distinct (r1, r2 & r3 or 2, 2i and -2i). So the homogeneous solution is:

y_H = Ae^(r1*x) + Be^(r2*x) + Ce^(r3*x)

wouldn't that be Ae^2x + Be^2ix + Ce^-2ix. .n that is different from the answer given by HallsofIvy. .??

 

[HallsofIvy]

wouldn't that be Ae^2x + Be^2ix + Ce^-2ix. .n that is different from the answer given by HallsofIvy. .??

But \(\displaystyle e^{2ix}= cos(2x)+ i sin(2x)\) and \(\displaystyle e^{-2ix}= cos(-2x)+ i sin(-2x)= cos(2x)- i sin(2x)\) because cosine is an "even" function and sine is an "odd" function. So \(\displaystyle Be^{2ix}+ Ce^{-2x}= Bcos(2x)+ Bi sin(2x)+ Ccos(2x)- Ci sin(2x)= (B+ C)cos(2x)+ (B- C)i sin(2x)\). Writing the "B" and "C" in my previous post as B' and C', we have B'= B+C and C'= (B- C)i.

(Besides, you were the one who said " for imaginary roots, y= e^ax [C₃ cos bx + C₄ ​sin bx]"!)

 

[Lost souls]

But \(\displaystyle e^{2ix}= cos(2x)+ i sin(2x)\) and \(\displaystyle e^{-2ix}= cos(-2x)+ i sin(-2x)= cos(2x)- i sin(2x)\) because cosine is an "even" function and sine is an "odd" function. So \(\displaystyle Be^{2ix}+ Ce^{-2x}= Bcos(2x)+ Bi sin(2x)+ Ccos(2x)- Ci sin(2x)= (B+ C)cos(2x)+ (B- C)i sin(2x)\). Writing the "B" and "C" in my previous post as B' and C', we have B'= B+C and C'= (B- C)i.

(Besides, you were the one who said " for imaginary roots, y= e^ax [C₃ cos bx + C₄ ​sin bx]"!)

yeah. .i realized as much. .euler's theorem. .thanx HallsofIvy and Subhotosh. .it's a great help. .