Section 1.4, problem 1
Find the general solution of:
xy' = x + y
We can re-arrange this to be:Using a change of variable such that u = y/x and y' = u + u'x we get:
u + u'x = 1 + u
With some re-arrangement we get:Integrating:
gives:
u = ln |x| + C
Changing back to y using y = ux, we get:
y = x (ln |x| +C)
Section 1.4, problem 2
Find the general solution of:
xy' - 2y = 3x
Re-arrangement gives:Using a change of variable such that u = y/x and y' = u + u'x we get:
u + u'x - 2u = 3
Re-arranging to separate the variables we have:Integrating both sides we get:
Solving:
ln |u+3| = ln |x| + C
Raising e to the power of both sides gives:
u + 3 = Cx
u = Cx - 3
Changing back to y using y = ux, we get:
y = Cx² - 3x
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