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Now, the roots of the characteristic equation is calculated as, ² +9=0

Now, the roots of the characteristic equation is calculated as, ² +9=0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Hello,

I have attached a picture im confused on how the values of A and B were found. Please explain.

The original DE is y''+9y=sin2x

Now, the roots of the characteristic equation is calculated as, ² +9=0 r = ±3i Therefore, the roots of the equations are r₁ = 3i and r₂ = -3i . The complementary function of the differential equation can be written as, Yc (x) = C₁ cos 3x + c₂ sin 3x From the given differential equation y" +9y = sin 2x, the value of f (x) = sin 2x . Therefore, the derivative of sin 2x involves the terms sin 2x, cos 2x and constant. The right hand side of the given differential equation is sin 2x. Consider the particular solution as yp (x) = A cos 2x + B sin 2x . Now differentiate yp (x) Ур A cos 2x + B sin 2x with respect to x. y'p (x) = = -2A sin 2x + 2B cos 2x Now differentiate y'p ( = y"p (x) = (x) = -2A sin 2x + 2B cos 2x with respect to x. = -4A Cos 2x – 4B sin 2x Now substitute the values of yp (x), y'p (x) and y" (x) in the differential equation y" +9y = sin 2x to obtain the Р P value of A. -4A Cos 2x - 4B sin 2x + 9 (A cos 2x + B sin 2x) 5A cos 2x + 5B sin 2x = sin 2x Compare the coefficient of cos 2x and sin 2x and constant to obtain the value of A. 5A = 0,5B = 1 Now solve the system of linear equation to obtain the value of A. 1 A = 0, B = ²/ = sin 2x
Transcribed Image Text:Now, the roots of the characteristic equation is calculated as, ² +9=0 r = ±3i Therefore, the roots of the equations are r₁ = 3i and r₂ = -3i . The complementary function of the differential equation can be written as, Yc (x) = C₁ cos 3x + c₂ sin 3x From the given differential equation y" +9y = sin 2x, the value of f (x) = sin 2x . Therefore, the derivative of sin 2x involves the terms sin 2x, cos 2x and constant. The right hand side of the given differential equation is sin 2x. Consider the particular solution as yp (x) = A cos 2x + B sin 2x . Now differentiate yp (x) Ур A cos 2x + B sin 2x with respect to x. y'p (x) = = -2A sin 2x + 2B cos 2x Now differentiate y'p ( = y"p (x) = (x) = -2A sin 2x + 2B cos 2x with respect to x. = -4A Cos 2x – 4B sin 2x Now substitute the values of yp (x), y'p (x) and y" (x) in the differential equation y" +9y = sin 2x to obtain the Р P value of A. -4A Cos 2x - 4B sin 2x + 9 (A cos 2x + B sin 2x) 5A cos 2x + 5B sin 2x = sin 2x Compare the coefficient of cos 2x and sin 2x and constant to obtain the value of A. 5A = 0,5B = 1 Now solve the system of linear equation to obtain the value of A. 1 A = 0, B = ²/ = sin 2x
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