Question #0c06b

2 Answers
Andrew I.
Jan 15, 2018

See below.

Explanation:

Solve this differential equation:

#dy/dx-1=0#

#->dy/dx=1#

#->intdy=intdx#

Solving for the general solution gives:

#y=x+C#

Where #C# is the constant of integration. As #C# is yet undetermined and can take any (complex) value we call this the general solution. #C# might be #+2# or #-3# (though without further information there is no reason it can't be something like #1,4,pi,sqrt(10)# or #1+i#). These two values of #C# would give us:

#y=x+2# or #y=x-3#

In these cases #C# has been determined so these are particular solutions. In other words they are just specific cases of the general solution. The value of #C# can usually be worked out when given "initial" or "boundary" conditions.

Jim H
Jan 15, 2018

Please see below.

Explanation:

For #y = x-3# we have #dy/dx = 1# so #dy/dx - 1 =0#

And also, for for #y = x+2# we have #dy/dx = 1# so #dy/dx - 1 =0#

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