Variation of Parameters

**MuhTheKuh** · January 14th 2012, 11:15 AM

I have a problem understanding a certain step using the Variation of Parameters Method.

take for example

y''+y=sec(x)tan(x)

Auxiliary Equation: m²+1=0 thus m= +/- 1

CF: Y1= Acos(x)+Bsin(x)

then I let y=A(x)cos(x)+B(x)sin(x)

and differentiate to

y'=-Asin(x)+Bcos(x)+[A'cos(x)+B'sin(x)] and set A'cos(x)+B'sin(x)=0 * (remember this for later)

differentiating again

y''= -Acos(x)-Bsin(x)-A'sin(x)+B'cos(x)

Substituting the yields now:

-A'sin(x)+B'cos(x)=sec(x)tan(x) ** (remember this for the next step)

so now I have to solve for A' and/or B'

I have no idea how to do this and when I checked the answer booklet it explained the next step by:

* cos(x) - ** sin(x): A'=tan²(x) = 1 - sec(x) A= x - tan(x) + C
and
* sin(x) - ** cos(x): B'=tan(x) B= ln[sec(x)] + D

I fully understand how to get up to the point were it says

* cos(x) - ** sin(x)

and I do understand how to integrate tan(x) etc, but I have no idea how I am supposed to get from my two marked (*) differentiations to A'= or B'=

I hope someone can help me as this is extremely frustrating and I have not been able to find anything that could help me, yet.

Thank you guys for your time, I appreciate it.

**alexmahone** · January 14th 2012, 11:47 AM

Originally Posted by MuhTheKuh

I have a problem understanding a certain step using the Variation of Parameters Method.

take for example

y''+y=sec(x)tan(x)

Auxiliary Equation: m²+1=0 thus m= +/- 1

CF: Y1= Acos(x)+Bsin(x)

then I let y=A(x)cos(x)+B(x)sin(x)

and differentiate to

y'=-Asin(x)+Bcos(x)+[A'cos(x)+B'sin(x)] and set A'cos(x)+B'sin(x)=0 * (remember this for later)

differentiating again

y''= -Acos(x)-Bsin(x)-A'sin(x)+B'cos(x)

Substituting the yields now:

-A'sin(x)+B'cos(x)=sec(x)tan(x) ** (remember this for the next step)

so now I have to solve for A' and/or B'

I have no idea how to do this and when I checked the answer booklet it explained the next step by:

* cos(x) - ** sin(x): A'=tan²(x) = 1 - sec(x) A= x - tan(x) + C
and
* sin(x) - ** cos(x): B'=tan(x) B= ln[sec(x)] + D

I fully understand how to get up to the point were it says

* cos(x) - ** sin(x)

and I do understand how to integrate tan(x) etc, but I have no idea how I am supposed to get from my two marked (*) differentiations to A'= or B'=

I hope someone can help me as this is extremely frustrating and I have not been able to find anything that could help me, yet.

Thank you guys for your time, I appreciate it.

These are your 2 equations:

$A'\cos x+B'\sin x=0$ *

$-A'\sin x+B'\cos x=\sec x\tan x$ **

In order to eliminate $B$ , we multiply the * equation by $\cos x$ , the ** equation by $\sin x$ and subtract to get

$A'(\cos^2x+\sin^2x)=-\sec x\tan x\sin x$

$A'=-\frac{\sin^2x}{\cos^2x}$

$=-\tan^2x$

$=1-\sec^2x$

In order to eliminate $A$ , we multiply the * equation by $\sin x$ , the ** equation by $\cos x$ and add to get

$B'(\sin^2x+\cos^2x)=\tan x$

$B'=\tan x$

**MuhTheKuh** · January 14th 2012, 11:57 AM

Thanks a lot, but somehow I did not know that -sec(x)tan(x)sin(x) are equal to -sin²(x)/cos²(x)
or that sec(x)tan(x)cos(x) = tan(x)
I'd hate to bother you (or anyone else) but is there something like a list where I can see that -sec(x)tan(x)sin(x)= ... etc?
or how someone would put it if i had to take
tan(x)cos(x)?

Thanks a lot!

**alexmahone** · January 14th 2012, 12:01 PM

Originally Posted by MuhTheKuh

Thanks a lot, but somehow I did not know that -sec(x)tan(x)sin(x) are equal to -sin²(x)/cos²(x)
or that sec(x)tan(x)cos(x) = tan(x)
I'd hate to bother you (or anyone else) but is there something like a list where I can see that -sec(x)tan(x)sin(x)= ... etc?
or how someone would put it if i had to take
tan(x)cos(x)?

Thanks a lot!

Just rewrite everything in terms of sines and cosines.

$-\sec x\tan x\sin x=-\frac{1}{\cos x}\frac{\sin x}{\cos x}\sin x=-\frac{\sin^2x}{\cos^2x}$

$\sec x\tan x\cos x=\frac{1}{\cos x}\tan x\cos x=\tan x$

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