Higher order differential equation

[Raghav Gupta]

Solve
y'' - 2y' + 2y = $e^x tanx$
What concept should we use if we know only solving first order differential equation?

 

[MisterX]

Some methods you could use are variation of parameters or a Fourier/Laplace/Green's function approach.

 

[DEvens]

Solve
y'' - 2y' + 2y = $e^x tanx$
What concept should we use if we know only solving first order differential equation?

If you know how to solve a first order equation, can you solve two coupled first order equations?

Let $v = y'$. Then convert your equation to this.

$v' - 2v + 2y = e^x \tan x$

Now you have two coupled first order equations.

 

[RUber]

Factor your differential operator.
$( \frac {d}{dx} - I ) ^2 y = e^x \tan x$
What sorts of functions solve the homogeneous problem?
$( \frac {d}{dx} - I ) ^2 y = 0$
Or...
Imagine that $y = f(x) e^x$ and take the derivatives. What do you find out about f(x)?

 

[pasmith]

Solve
y'' - 2y' + 2y = $e^x tanx$
What concept should we use if we know only solving first order differential equation?

Factor your differential operator.
$( \frac {d}{dx} - I ) ^2 y = e^x \tan x$

Unfortunately \lambda^2 - 2\lambda + 2 = (\lambda - 1)^2 +1 does not factorise over the reals (which isn't really a problem but not ideal for someone not familiar with second-order linear differential operators with constant coefficients).

That said, I would set y(x) = f(x)e^{kx} and choose k to eliminate the f term. This eventually yields a first-order ODE for f', and most of the resulting integrals can be done analytically.

 

[RUber]

Unfortunately \lambda^2 - 2\lambda + 2 = (\lambda - 1)^2 +1 does not factorise over the reals (which isn't really a problem but not ideal for someone not familiar with second-order linear differential operators with constant coefficients).

Thanks pasmith. Good catch.

 

[epenguin]

Can you tell us where this problem comes from?
Doesn't even to solve y''= tan x require special functions that would not be considered elementary as you are essentially requiring?
Unless there is something special about your equation allowing some clever way round. (I cannot see that #4 is it.)

 

[HallsofIvy]

If all you know is "solving first order equations" then you can't solve this equation. It simply cannot be reduced to a first order equation! There is a reason for learning how to solve higher order equations!

 

[Raghav Gupta]

Read and understood the concept of second order linear non-homogeneous differential equation.
Here we have to find first the complementary function and then the particular integral
y=y_c+y_p
So we have the characteristic or auxiliary equation first as m²-2m+2=0
I know solving complementary function
m=1± i
y_c=e^x(cosx+sinx)
Now for y_p I know the method of undetermined coefficients.
For e^x we can simply substitute y= Ae^x but here the R.H.S is f(x)e^x.
What to do?

 

[epenguin]

OK so you now know it seems some of the method for non-homogeneous (constant coefficient) linear d.e's. and have solved for the complementary function.

In the most widely needed applications of these x is time, and the RHS, called the 'forcing function', is commonly sine/cosine, or several such, or exponential or combination. These are daily bread of electrical or radio engineers etc. There are some widely taught methods quite adequate for dealing with this kind of function, which work because these functions have nice properties, namely that all their derivatives are also sin/cosine/exponential.

The function you're given, with tan, unfortunately does not have quite such nice properties, so you are trying to run before you can walk. (A very good thing IMO! ). It is also rather unphysical if x is time, as it is periodic but going to infinity in finite time (x).

I think I can make one of the cobbling methods work for this. Maybe I wil come back on this. However you do not want to waste time on methods that are hard to puzzle out so I think you can actually use the variation of parameters method you mentioned. In Dr. Paul's Online Math Notes you can find not only a back-up explanation of the method but a worked example that is almost the same as your problem.

The second sentence of my #7 was a misconception btw.

 

[HallsofIvy]

The "basic" solution for a linear, homogeneous, differential equation with constant coefficients is of the form e^{\alpha x} where \alpha is a solution to the associated characteristic equation. From that we also get sine and cosine solutions (if \alpha is complex) and powers of x (if \alpha is a multiple root). The method of "undetermined coefficients" only works if the right hand side of a non-homogeneous equation is one of those types of function.

What you need to use here is called "variation of parameters", that MisterX mentioned before. If u(x) and v(x) are two independent solutions to the homogeneous equation, then a general solution is of the form "Cu(x)+ Dv(x)" where C and D are constants. Those are the "parameters" we are going to vary. Look for solutions of the form p(x)u(x)+ q(x)v(x), with p and q functions of x. With y(x)= p(x)u(x)+ q(x)v(x), y'= p'u+ pu'+ q'v+ qv'. Because there are an infinite number of possible solutions, there are an infinite number of possible "p" and "q". We "narrow the search", and simplify the problem, by searching for solutions such that p'u+ q'v= 0. That leaves y'= pu'+qv' and the second derivative is y''= p'u'+ pu''+ q'v'+ qv''. Now, put those y, y', and y'' into the differential equation.

Because u and v satisfy the homogeneous equation, all terms involving only p and q (without being differentiated) will cancel. Since we required that p'u+ q'v= 0 there were be no p'' or q''. That is, we will have an equation involving only p' and q'. That, together with p'u+ q'v= 0, give us two equations to solve, algebraically, for p' and q'.