If the ODE y\'\' + 20y\' + 36y = tan x is solved by the variation of parameters
ID: 2969063 • Letter: I
Question
If the ODE y'' + 20y' + 36y = tan x is solved by the variation of parameters technique, with A(x) and B(x) representing the unknown functions, then a pair of linear equations in A' and B' will be Question 5 options: A) A'e3x + B'e-12x = 03A'e3x - 12B'e-12x = tan x B) A'e3x + B'e12x = 0
3A'e3x + 12B'e12x = tan x C) A'e-3x + B'e12x = 0
-3A'e-3x + 12B'e12x = tan x D) A'e-3x + B'e-12x = 0
-3A'e-3x - 12B'e-12x = tan x E) A'e4x + B'e-9x = 0
4A'e4x - 9B'e-9x = tan x F) A'e4x + B'e9x = 0
4A'e4x + 9B'e9x = tan x G) A'e-4x + B'e9x = 0
-4A'e-4x + 9B'e9x = tan x H) A'e-4x + B'e-9x = 0
-4A'e-4x - 9B'e-9x = tan x I) A'e2x + B'e18x = 0
2A'e2x + 18B'e18x = tan x J) A'e-2x + B'e18x = 0
-2A'e-2x + 18B'e18x = tan x K) A'e-2x + B'e-18x = 0
-2A'e-2x - 18B'e-18x = tan x L) A'e2x + B'e-18x = 0
2A'e2x - 18B'e-18x = tan x M) none of these is possible If the ODE y'' + 20y' + 36y = tan x is solved by the variation of parameters technique, with A(x) and B(x) representing the unknown functions, then a pair of linear equations in A' and B' will be If the ODE y'' + 20y' + 36y = tan x is solved by the variation of parameters technique, with A(x) and B(x) representing the unknown functions, then a pair of linear equations in A' and B' will be A) A'e3x + B'e-12x = 0
3A'e3x - 12B'e-12x = tan x B) A'e3x + B'e12x = 0
3A'e3x + 12B'e12x = tan x C) A'e-3x + B'e12x = 0
-3A'e-3x + 12B'e12x = tan x D) A'e-3x + B'e-12x = 0
-3A'e-3x - 12B'e-12x = tan x E) A'e4x + B'e-9x = 0
4A'e4x - 9B'e-9x = tan x F) A'e4x + B'e9x = 0
4A'e4x + 9B'e9x = tan x G) A'e-4x + B'e9x = 0
-4A'e-4x + 9B'e9x = tan x H) A'e-4x + B'e-9x = 0
-4A'e-4x - 9B'e-9x = tan x I) A'e2x + B'e18x = 0
2A'e2x + 18B'e18x = tan x J) A'e-2x + B'e18x = 0
-2A'e-2x + 18B'e18x = tan x K) A'e-2x + B'e-18x = 0
-2A'e-2x - 18B'e-18x = tan x L) A'e2x + B'e-18x = 0
2A'e2x - 18B'e-18x = tan x M) none of these is possible A) A'e3x + B'e-12x = 0
3A'e3x - 12B'e-12x = tan x B) A'e3x + B'e12x = 0
3A'e3x + 12B'e12x = tan x C) A'e-3x + B'e12x = 0
-3A'e-3x + 12B'e12x = tan x D) A'e-3x + B'e-12x = 0
-3A'e-3x - 12B'e-12x = tan x E) A'e4x + B'e-9x = 0
4A'e4x - 9B'e-9x = tan x F) A'e4x + B'e9x = 0
4A'e4x + 9B'e9x = tan x G) A'e-4x + B'e9x = 0
-4A'e-4x + 9B'e9x = tan x H) A'e-4x + B'e-9x = 0
-4A'e-4x - 9B'e-9x = tan x I) A'e2x + B'e18x = 0
2A'e2x + 18B'e18x = tan x J) A'e-2x + B'e18x = 0
-2A'e-2x + 18B'e18x = tan x K) A'e-2x + B'e-18x = 0
-2A'e-2x - 18B'e-18x = tan x L) A'e2x + B'e-18x = 0
2A'e2x - 18B'e-18x = tan x M) none of these is possible A) A'e3x + B'e-12x = 0
3A'e3x - 12B'e-12x = tan x B) A'e3x + B'e12x = 0
3A'e3x + 12B'e12x = tan x C) A'e-3x + B'e12x = 0
-3A'e-3x + 12B'e12x = tan x D) A'e-3x + B'e-12x = 0
-3A'e-3x - 12B'e-12x = tan x E) A'e4x + B'e-9x = 0
4A'e4x - 9B'e-9x = tan x F) A'e4x + B'e9x = 0
4A'e4x + 9B'e9x = tan x G) A'e-4x + B'e9x = 0
-4A'e-4x + 9B'e9x = tan x H) A'e-4x + B'e-9x = 0
-4A'e-4x - 9B'e-9x = tan x I) A'e2x + B'e18x = 0
2A'e2x + 18B'e18x = tan x J) A'e-2x + B'e18x = 0
-2A'e-2x + 18B'e18x = tan x K) A'e-2x + B'e-18x = 0
-2A'e-2x - 18B'e-18x = tan x L) A'e2x + B'e-18x = 0
2A'e2x - 18B'e-18x = tan x M) none of these is possible
Explanation / Answer
characteristic equation:
r^2 + 20r + 36 = 0
(r+2)(r+18) = 0
-> roots are -2 , -18 -> y1 = e^(-2x) , y2 = e^(-18x)
-> A'y1 + B'y2 = 0 -> A'.e^(-2x) + B'.e^(-18x) = 0
A'y1' + B'y2' = tan(x) -> -2A'.e^(-2x) -18 B'.e^(-18x) = tan(x)
answer K.
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