How Do You Solve This Trigonometric Differential Equation from Calculus II?

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SUMMARY

The discussion focuses on solving the trigonometric differential equation defined by the expression \( y = \sin x \cos x - \cos^2 x \) and the equation \( 2y + y' = 2\sin(2x) - 1 \) with the initial condition \( y(\pi/4) = 0 \). The user, Ed, attempts to simplify the equation but struggles to prove the equality \( \sin(2x) + 2\sin(x) - 1 = 2\sin(2x) - 1 \). A key insight reveals that the equality is incorrect, leading to the conclusion that Ed likely made an error in his calculations prior to this step.

PREREQUISITES
  • Understanding of trigonometric identities, specifically the double angle formulas.
  • Familiarity with differential equations and their solutions.
  • Knowledge of calculus concepts, particularly derivatives and initial conditions.
  • Ability to manipulate algebraic expressions involving trigonometric functions.
NEXT STEPS
  • Review the double angle formulas for sine and cosine.
  • Practice solving trigonometric differential equations using initial conditions.
  • Learn about the implications of algebraic manipulation in trigonometric identities.
  • Explore common mistakes in calculus, particularly in solving differential equations.
USEFUL FOR

Students enrolled in Calculus II, particularly those struggling with trigonometric differential equations, as well as educators looking for examples of common pitfalls in calculus problem-solving.

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Homework Statement


y=sinxcosx-cos^{2}x Differential equation= 2y+y'=2sin(2x)-1
Initial condition y(pi/4)=0






The Attempt at a Solution



2y=2sinxcosx-2cos^{2}x

y'=-sin^{2}x+cos^{2}x+2cosxsinx

ok... using double angle formulas... I get
2y=Sin2x-2cos^{2}x

y'=2cos^{2}x-1+2sinx

2y+y'
(Sin2x-2Cos^{2}x)+(2cos^{2}x-1+2sinx)
the 2cos mess cancells out leaves you with

Sin2x + (-1) +2Sinx
I don't understand how that equals 2sin2x-1
as the book says it does and the solution shows.

Stupid question I know... I struggle with the easy stuff.

Thanks
-Ed
 
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You're trying to prove that sin(2x) + 2sin(x)-1 = 2sin(2x)-1

Edit: I have to leave in a hurry, but this equality is not true.
In the RHS, 2sin(2x)-1=sin(2x)+sin(2x)-1

\implies sin(2x) + 2sin(x)-1 = sin(2x)+sin(2x)-1
Therefore, cancelling the sin(2x)-1's from the LHS and RHS,
2sin(x) \neq sin(2x) since by the double angle formula,

sin(2x) = 2sin(x)cos(x)

So, you probably made a mistake somewhere before.
 
Last edited:
awesome... I'm totally screwed for tomorrow. lol

Cal 2 during the summer was a terrible idea.
 

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