Variation of parameters method

[djeitnstine]

Homework Statement

y''+y=tan(x)+e^{3x}-1

Homework Equations

homogeneous solution:
y_{hom..}=C_{1}cos(x)+C_{2}sin(x)

particular solution:
y_{parti..}=v_{1}' cos(x)+v_{2}' sin(x)

The Attempt at a Solution

v_{1}' cos(x)+v_{2}' sin(x)=0 (1)

-v_{1}' sin(x)+v_{2}' cos(x) = tan(x)+e^{3x}-1 (2)

(1)* by sin (2)* by cos

and solve for v2

sin(x)+e^{3x}cos(x)-cos(x)=v_{2}'

-cos(x)+ \frac{3}{10}e^{3x}cos(x)-\frac{1}{10}sin(x)-sin(x)=v_{2}

Now finding v1 is the problem here. if I reverse the multiplication and do (1)* by cos (2)* by sin it seems even worse =(. I saw the complete answer and if I remember correctly its y_{p}=\frac{1}{2}cos(x)ln(sin(x)+1)-\frac{1}{2}cos(x)ln(sin(x)-1)-\frac{1}{10}e^{3x}+1

I really just need to know what to multiply by to get going

 

[gabbagabbahey]

particular solution:
y_{parti..}=v_{1}' cos(x)+v_{2}' sin(x)

Why the primes? Why not just use y_p=v_{1}\cos(x)+v_{2}\sin(x) as your ansatz (assumed form)?

v_{1}' cos(x)+v_{2}' sin(x)=0 (1)

-v_{1}' sin(x)+v_{2}' cos(x) = tan(x)+e^{3x}-1 (2)

Where are these equations coming from?

Your particular solution should satisfy y_p''+y_p=\tan(x)+e^{3x}-1 not y_p=0 or (2)

 

[HallsofIvy]

As gabbagabbahey said, drop the primes from you solution- they are confusing.
y(x)= v1(x)cos(x)+ v2(x)sin(x)
Differenitate:
y'= v1' cos(x)- v1 sin(x)+ v2' sin(x)+ v2 cos(x)
Now, we can simplify this by "restricting" our search. Out of the infinitely many v1 and v2 that would satisfy the equation, we only look for those such that
v1' cos(x)+ v2' sin(x)= 0

So now
y'= -v1 sin(x)+ v2 cos(x)
Differentiate again: y"= -v1' sin(x)- v1 cos(x)+ v2' cos(x)- v2 sin(x).

Put that and the equation for y' into the original differential equation and it gives one equation for v1' and v2' (all of the v1 and v2 terms without ' will cancel). That, together with v1' cos(x)+ v2' sin(x)= 0 gives to equations to solve (algebraically) for v1' and v2' and then you integrate to find v1 and v2.

 

[djeitnstine]

Ok thanks I finally got the answer.

 

[djeitnstine]

Why the primes? Why not just use y_p=v_{1}\cos(x)+v_{2}\sin(x) as your ansatz (assumed form)?



Where are these equations coming from?

Your particular solution should satisfy y_p''+y_p=\tan(x)+e^{3x}-1 not y_p=0 or (2)

These equations come from a proof shown by various texts including my professor. One example is at http://tutorial.math.lamar.edu/ under the Diff Eq. pdf if you'd like to see.