Differential Equation

[car0le_la]

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.

 

[Steven G]

When you square sin(x) + cos(x) you do not get sin²(x) + cos²(x). You are in differential equations not basic algebra--you should know by now that (a+b)² is not a²+ b²

Also if y²=1 it does NOT follow that y=sqrt(1) which is better known as 1. That is if y²=1 it does NOT follow that y= 1.

Hint: 1st compute dy/dx. Then add that to y and see if you get 2sin(x) or 2cos(x). That will help with choice 1 and 2. Then compute dy/dx - y and see if you get -2sin(x).

 

[car0le_la]

Yeah, I should have realized that error: (a+b)² does not equate to ²+b²
?
...
dy/dx =cosx-sinx
y+cosx-sinx=2sinx
y=3sinx-cosx
so this means that it isn´t choice 1
...
y+cosx-sinx=2cosx
y=cosx+sinx
so choice 2 is good
...
cosx-sinx-y=-2sinx
-cosx+sinx+y=2sinx
y=2sinx-sinx+cosx
y=sinx+cosx
so choice 3 is good
...
which would fit with the answer being D) ii. and iii.
Thanks!