Variation of Parameters, system of equations

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The discussion revolves around solving the differential equation y'' + 25y = cot(5x) using the method of variation of parameters. The user correctly identifies the complementary solution y_h = Acos(5x) + Bsin(5x) and the need to find a particular solution y_p. They express confusion about solving the resulting system of equations derived from substituting u_1 and u_2 into the equations. Suggestions include substituting u_1' = -u_2'tan(5x) into the equations and considering the Wronskian for simplification. The user seeks clarity on the algebraic steps necessary to find the particular solution.
Linday12
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Homework Statement


y''+25y=cot(5x)
Find one possible solution


The Attempt at a Solution


I don't have any background in linear algebra, so I can't use cramers rule as a heads up, so I have to solve the system of equations (no linear algebra for this course is needed).

Ok, so I take the auxiliary, r^2=-25, r=+/-5i

y_h=Acos(5x)+Bsin(5x)

y_1=cos(5x) y_2=sin(5x)

Then,
u_1'cos(5x)+u_2'sin(5x) = 0
-5u_1'sin(5x)+5u_2'cos(5x) = cot(5x)

I am stumped at how to solve this system of equations. I think the rest before it is right..?

Seems like the algebra has got me.
 
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Perhaps I'm going about this completely wrong then. How would I start out solving y''+25y=cot(5x).

(I should be more specfic, when I say solve, I mean to find A solution, so y_h and y_p. I'm just not sure about how to find the y_p=u1(y_1)+u2(y_2))
 
Linday12 said:
Ok, so I take the auxiliary, r^2=-25, r=+/-5i

y_h=Acos(5x)+Bsin(5x)

y_1=cos(5x) y_2=sin(5x)

Then,
u_1'cos(5x)+u_2'sin(5x) = 0
-5u_1'sin(5x)+5u_2'cos(5x) = cot(5x)

I am stumped at how to solve this system of equations. I think the rest before it is right..?

Seems like the algebra has got me.

u1' = -u2'tan(5x), sub that into the other equation.

this may help you as well sin2x+cos2x=1

Though, you could just use the direct result to make this easier. You are sure you can't use the Wronskian?
 
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