Variation of Parameters, system of equations

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SUMMARY

The discussion centers on solving the differential equation y'' + 25y = cot(5x) using the method of variation of parameters. The auxiliary equation yields complex roots, leading to the homogeneous solution y_h = Acos(5x) + Bsin(5x). The user struggles with the system of equations derived from the variation of parameters, specifically in finding the particular solution y_p using the functions y_1 = cos(5x) and y_2 = sin(5x). The conversation highlights the need for clarity in applying the method and suggests using the Wronskian for simplification.

PREREQUISITES
  • Understanding of second-order linear differential equations
  • Familiarity with the method of variation of parameters
  • Knowledge of trigonometric identities, specifically sin²x + cos²x = 1
  • Basic concepts of complex numbers and their applications in differential equations
NEXT STEPS
  • Study the method of variation of parameters in detail
  • Learn how to compute the Wronskian for a set of functions
  • Practice solving second-order linear differential equations with non-homogeneous terms
  • Explore the implications of using trigonometric identities in solving differential equations
USEFUL FOR

Students studying differential equations, educators teaching linear algebra concepts, and anyone seeking to enhance their problem-solving skills in applied mathematics.

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?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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