Differential EquationHard
Question
If y1(x) and y2(x) are two solutions of 
+ f(x) y = r(x) then y1(x) + y2(x) is solution of :
+ f(x) y = r(x) then y1(x) + y2(x) is solution of :Options
A.
+ f(x) y = 0
+ f(x) y = 0B. + 2f(x) y = r(x)
+ 2f(x) y = r(x)C. + f(x) y = 2 r(x)
+ f(x) y = 2 r(x)D. + 2f (x) y = 2r(x)
+ 2f (x) y = 2r(x)Solution
+ fy1 = r
+ fy2 = rAdd
(y1 + y2) + f(y1 + y2) = 2rhere
+ f(x)y = 2rCreate a free account to view solution
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