Having a problem trying to find the appropriate guess (Yp) for the following O.D

Question

Having a problem trying to find the appropriate guess (Yp) for the following O.D.E

Y'' - 2Y' + 3Y = [t*cos(2t)] + [e^t * sin(sqrt(2)*t)]

If you can please just write the approximate Yp guesses and their derivatives so I can compare my work, and let me know how you got them I'd appreciate it. I have the book, I just can't seem to find a problem like this one with an explanation. I've already found Yc, and I would like to do the work myself, just having a problem with Yp. Must use the Method of Undetermined Coefficients.

Explanation / Answer

Let's jump straight to an example;

y'' + 4y = 3csc(t)

Noting that the corresponding homogeneous equation is;

y'' + 4y = 0

We first solve this equation.

r12 = √(-4(1)(4))/2(1) = 2i

remember that solution will be in form;

eλt[cos(μt) + sin(μt)]

where in this case λ = 0 and μ = 2, so

yc(t) = c1cos(2t) + c2sin(2t)

The basic idea is to replace the constance c1 and c2 with functions u1(t) and u2(t) and solve for these functions.

Starting with the equation;

y = u1(t)cos(2t) + u2(t)sin(2t)

we can differentiate to optain;

y' = u'1(t)cos(2t) + u'2(t)sin(2t) - 2u1(t)sin(2t) + 2u2(t)cos(2t)

Since we only have one initial condition so far, yet two unknown variables, this would give us infinite many solutions. Let us impose a second condition so that we have one final solution. Here it is not important why we can do this;

We require that;

u'1(t)cos(2t) + u'2(t)sin(2t) = 0, so;

y' = 2u2(t)cos(2t) - 2u1(t)sin(2t)

y'' = 2u'2(t)cos(2t) - 2u'1(t)sin(2t) - 4u2(t)sin(2t) - 4u1(t)cos(2t)

Substitude these equations back into the original equation;

[2u'2(t)cos(2t) - 2u'1(t)sin(2t) - 4u2(t)sin(2t) - 4u1(t)cos(2t)] + 4[u1(t)cos(2t) + u2(t)sin(2t)] = 3csc(t)

2u'2(t)cos(2t) - 2u'1(t)sin(2t) = 3csc(t)


From our second set condition;
u'1(t)cos(2t) + u'2(t)sin(2t) = 0

u'2(t) = -u'1(t)cos(2t)/sin(2t)


Substitude;
2[-u'1(t)cos(2t)/sin(2t)]cos(2t) - 2u'1(t)sin(2t) = 3csc(t)

Simplify;

u'1(t) = -(3csc(t)sin(2t))/2 = -3cos(t)

Substituging once more;

u'2(t) = -u'1(t)cos(2t)/sin(2t)
u'2(t) = [-3cos(t)]cos(2t)/sin(2t)
u'2(t) = (3/2)csc(t) -3sin(t)

Now that we have optained u'1(t) and u'2(t), Integrate;

u1(t) = -3sin(T) + c1
u2(t) = (3/2)ln|csc(t) - cot(t)| + 3cot(t) + c2(t)

Finally, substitude u1(t) and u2(t) into the y expression;

y = [-3sin(T) + c1]cos(2t) + [(3/2)ln|csc(t) - cot(t)| + 3cot(t) + c2(t)]sin(2t)


That probably looked more confusing than need be, so lets look at an arbitrary function to see a set by set method and prove this can be used for any Second Order Linear Nonhomogeneous Equation.

Let us start with the general equation;
y'' + p(t)y' + q(t)y = g(t)

The general solution to the corresponding homogeneous equation will be;
yc(t) = c1y1(t) + c2y2(t)

This is from the assumption that the equation has constant coefficients. Now in the general solution, replace constants with functions u.

y = u1(t)y1(t) + u2(t)y2(t)

Take the derivative;
y' = u'1(t)y1(t) + u'2(t)y2(t) + u1(t)y'1(t) + u2(t)y'2(t)

For a second condition set terms with u' equal to zero;
u'1(t)y1(t) + u'2(t)y2(t) = 0

This gives;
y' = u1(t)y'1(t) + u2(t)y'2(t)

Differentiate again and plug y, y', and y'' into the original equation;

y'' = u1(t)y''1(t) + u2(t)y''2(t) + u'1(t)y'1(t) + u'2(t)y'2(t)

[u1(t)y''1(t) + u2(t)y''2(t) + u'1(t)y'1(t) + u'2(t)y'2(t)] + p(t)[u1(t)y'1(t) + u2(t)y'2(t)] + q(t)[u1(t)y1(t) + u2(t)y2(t)] = g(t)

Rearranging;
u1(t)[y''1(t) + p(t)y'1(t) + q(t)y1(t)] + u2(t)[y''2(t) + p(t)y'2(t) + q(t)y2(t)] + u'1(t)y'1(t) + u'2(t)y'2(t)] = g(t)

Since both y1 and y2 are solutions to the corresponding homogeneous equation, the expressions in brackets equal zero, leaving;

u'1(t)y'1(t) + u'2(t)y'2(t) = g(t)

Using this equation and the previous equation;
u'1(t)y'1(t) + u'2(t)y'2(t) = 0

substituation can be used and integration can be done to find u1and u2.

u'1(t) = -y2(t)g(t)/W(y1,y2)(t)

u'2(t) = y1(t)g(t)/W(y1,y2)(t)

u1(t) = -∫(y2(t)g(t)/W(y1,y2)(t))dt + c1

u2(t) = ∫(y1(t)g(t)/W(y1,y2)(t))dt + c2

Where then;
Y(t) =
-y1(t)∫(y2(t)g(t)/W(y1,y2)(t))dt + y2∫(y1(t)g(t)/W(y1,y2)(t))dt

and the general solution is;
y = c1y1(t) + c2y2(t) + Y(t)

I realize this is a little confusing to follow, so let me sum up what you really need to know without deriving everything everytime.

First;
You must find the solutions y1 and y2 of the homogeneous equation.

Then use these two formulas:

u'1y1 + u'2y2 = 0
u'1y'1 + u'2y'2 = g(t)

If there is a term infront of the y'' term, it must be divided out to give the correct g(t).

From this system of equations, where y and y' are known, u'1 and u'2 can be found, then integrated.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.