Variation of Parameters (1 Viewer)

addikaye03

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1) y''+y=cotx
2) y''+y=cosecx

Can someone talk me through the solutions, only new to the method.

Cheers

PS: So it's based on finding: y(x)=yc+yp where yc is the complimentary solution and yp is the particular. This falls under 2 order non-homogenuous differential equations.
 
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Iruka

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OK, I'll just go through the algorithm. You look up why it works later.

Firstly, you solve the homogeneous equation. In this case, you get two fundamental solutions, u1=cos x and u2=sin x.

Next you find the Wronskian of the these solutions, W. (In this case it is 1, so not particularly interesting.)

Then yp=u1*v1+ u2*v2, where u1, and u2 are as above, and

v1= - int u2*phi(x)/W dx

v2=int u1phi(x)/W dx

where phi(x) = RHS of the inhomogeneous equation. In the case of your first example, you solve

v1= - int sin(x)*cot(x) dx and

v2=int cos(x)*cot(x) dx.

As a result, you get

v1= - sin(x) and v2= cos(x) + ln(csc(x)- cot(x)).

Putting this altogether, we have the full solution

y(x) = -cos(x)*sin(x)+sin(x)*(cos(x)+ln(csc(x)-cot(x))) + a*cos(x) + b*sin(x), for some arbitrary constants a and b.
 

addikaye03

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OK, I'll just go through the algorithm. You look up why it works later.

Firstly, you solve the homogeneous equation. In this case, you get two fundamental solutions, u1=cos x and u2=sin x.

Next you find the Wronskian of the these solutions, W. (In this case it is 1, so not particularly interesting.)

Then yp=u1*v1+ u2*v2, where u1, and u2 are as above, and

v1= - int u2*phi(x)/W dx

v2=int u1phi(x)/W dx

where phi(x) = RHS of the inhomogeneous equation. In the case of your first example, you solve

v1= - int sin(x)*cot(x) dx and

v2=int cos(x)*cot(x) dx.

As a result, you get

v1= - sin(x) and v2= cos(x) + ln(csc(x)- cot(x)).

Putting this altogether, we have the full solution

y(x) = -cos(x)*sin(x)+sin(x)*(cos(x)+ln(csc(x)-cot(x))) + a*cos(x) + b*sin(x), for some arbitrary constants a and b.
Thanks alot, that really helps. They seem like quite easy questions if you understand the method. Thanks again mate
 

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