The function y=sinx+cosx is a solution of which differential equation?
i. y + dy = 2 sinx
dx
ii. y + dy = 2 cosx
dx
iii. dy - y = -2 sinx
dx
a) i only
b) ii only
c) iii only
d) ii and iii only
My work:
Attempt 1
y=sinx +cosx
dy = cosx-sinx
dx
Attempt 2:
y=sinx+cosx
y²=sin²x+cos²x
y²=1 because sin²x+cos²x=1
but then I would be stuck with y=√1 and then y´=0
I dont get how to get to any of the options of the three roman numerals. Especially the part where the roman numerals are able to maintain the y depite being the differential equation. And from where does that addtional 2 value appear? The answer says it's D but I don't know how they got to that.
i. y + dy = 2 sinx
dx
ii. y + dy = 2 cosx
dx
iii. dy - y = -2 sinx
dx
a) i only
b) ii only
c) iii only
d) ii and iii only
My work:
Attempt 1
y=sinx +cosx
dy = cosx-sinx
dx
Attempt 2:
y=sinx+cosx
y²=sin²x+cos²x
y²=1 because sin²x+cos²x=1
but then I would be stuck with y=√1 and then y´=0
I dont get how to get to any of the options of the three roman numerals. Especially the part where the roman numerals are able to maintain the y depite being the differential equation. And from where does that addtional 2 value appear? The answer says it's D but I don't know how they got to that.
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