How Can Variation of Parameters Solve Complex ODEs?

[brad sue]

Hi ,please I need help to find the solution of those equations:

y''+ 4y'=tan(2x)

y'' -6y' +9y=e^3x/x

My problem is to find the model for the particular solutions..
Can someone help me ??

 

[HallsofIvy]

You can't. The "method of undetermined coefficients" only works when the right hand side is one of the kinds of functions we expect as solutions to a homogeneous linear equation with constant coefficients: exponentials, sine or cosine, polynomials, and combinations of those. tan(2x) and e^3x/ x are not of that type.

Looks to me like you moved into the section on "variation of parameters" in your textbook! I'll walk you through the first one:

The associated homogeneous equation is y"+ 4y'= 0 which has characteristic equation r²+ 4r= r(r+4)= 0 and characteristic values r= 0, r= -4. The general solution to the homogeneous equation is y= C+ De^-4x (Ce^0x= C, of course.)

To find a specific solution to the full equation we look for a solution of the type y(x)= u(x)+ v(x)e^-4x. Do you see why it's called "variation of parameters"- the two constants in the solution (parameters) are allowed to vary- they are functions of x now.

That's no restriction on y at all. In fact, there are an infinite number of ways to do that- for any y(x) whatsoever, pick any function of u(x) and solve for v(x).

Differentiate y: y'= u'+ v'e^-4x- 4ve^-4x
Now assume that u'+ v'e^-4x= 0, Remember that there are an infinite number of u and v that will work. This just restricts those we are willing to use- and simplifies the problem enormously.

Since u'+ v'e^-4x= 0, there is no u' or v' :
y'= -4ve^-4x. Differentiate again:
y"= -4v'e^-4x+ 16ve^-4x.

Now put those into the differential equation:
y"+ 4y'= -4v'e^-4x+ 16ve^-4x-16ve^-4x= -4v'e^-4x= tan(2x).

Now we have two equations for u' and v': u'+ v'e^-4x= 0 and
4v'e^-4x= tan(2x). In this simple case, they are "partially uncoupled". We can solve the second equation immediately for v':
v'= (1/4)e^4xtan(2x) and put that into the first equation to solve for u': u'= -(1/4)tan(2x). To find u and v, just integrate!

There's the rub![/sub] Except for the simple cases where "undetermined coefficients" would also work, generally you wind up with really wicked integrals! What is commonly done is to leave the answers in integral form:
$$u(x)= -(1/4) \int_0^x tan(t/2)dx$$
and
$$v(x)= (1/4) \int_0^x e^{4t}tan(2t) dt$$
I can set those up as "definite" integrals since I am only seeking specific solutions, not the general solution. Also I am using t as the variable of integration inside the integrals rather than x because I want to make it clear that the exponentials in
$$e^{-4x}\int_0^x e^{4t}tan(2t) dt$$
do NOT cancel.

The general solution to the differential equation is
$$y(x)= C+ De{-4x}+ -(1/4) \int_0^x tan(t/2)dx+ (1/4)e^{-4x}\int_0^x e{4t}tan(2t) dt$$

 

[brad sue]

You can't. The "method of undetermined coefficients" only works when the right hand side is one of the kinds of functions we expect as solutions to a homogeneous linear equation with constant coefficients: exponentials, sine or cosine, polynomials, and combinations of those. tan(2x) and e^3x/ x are not of that type.

Looks to me like you moved into the section on "variation of parameters" in your textbook! I'll walk you through the first one:

The associated homogeneous equation is y"+ 4y'= 0 which has characteristic equation r²+ 4r= r(r+4)= 0 and characteristic values r= 0, r= -4. The general solution to the homogeneous equation is y= C+ De^-4x (Ce^0x= C, of course.)

To find a specific solution to the full equation we look for a solution of the type y(x)= u(x)+ v(x)e^-4x. Do you see why it's called "variation of parameters"- the two constants in the solution (parameters) are allowed to vary- they are functions of x now.

That's no restriction on y at all. In fact, there are an infinite number of ways to do that- for any y(x) whatsoever, pick any function of u(x) and solve for v(x).

Differentiate y: y'= u'+ v'e^-4x- 4ve^-4x
Now assume that u'+ v'e^-4x= 0, Remember that there are an infinite number of u and v that will work. This just restricts those we are willing to use- and simplifies the problem enormously.

Since u'+ v'e^-4x= 0, there is no u' or v' :
y'= -4ve^-4x. Differentiate again:
y"= -4v'e^-4x+ 16ve^-4x.

Now put those into the differential equation:
y"+ 4y'= -4v'e^-4x+ 16ve^-4x-16ve^-4x= -4v'e^-4x= tan(2x).

Now we have two equations for u' and v': u'+ v'e^-4x= 0 and
4v'e^-4x= tan(2x). In this simple case, they are "partially uncoupled". We can solve the second equation immediately for v':
v'= (1/4)e^4xtan(2x) and put that into the first equation to solve for u': u'= -(1/4)tan(2x). To find u and v, just integrate!

There's the rub![/sub] Except for the simple cases where "undetermined coefficients" would also work, generally you wind up with really wicked integrals! What is commonly done is to leave the answers in integral form:
$$u(x)= -(1/4) \int_0^x tan(t/2)dx$$
and
$$v(x)= (1/4) \int_0^x e^{4t}tan(2t) dt$$
I can set those up as "definite" integrals since I am only seeking specific solutions, not the general solution. Also I am using t as the variable of integration inside the integrals rather than x because I want to make it clear that the exponentials in
$$e^{-4x}\int_0^x e^{4t}tan(2t) dt$$
do NOT cancel.

The general solution to the differential equation is
$$y(x)= C+ De{-4x}+ -(1/4) \int_0^x tan(t/2)dx+ (1/4)e^{-4x}\int_0^x e{4t}tan(2t) dt$$

THANK YOU VERY MUCH for those precise explanations.
I work the second one and it was OK...
B