Answer Key for Quiz 6 (section A)
1. If y=A+B
x+C
x2=A+Bx−1+C x−2, then
dy
dx = 0 + B(−1)x−2+C(−2)x−3=−B
x2−2C
x3.
Substituting these expressions for yand dy
dx into the differential equation we get
1 = x4dy
dx +x2y
=x4µ−B
x2−2C
x3¶+x2µA+B
x+C
x2¶
=−Bx2−2C x +Ax2+Bx +C
= (A−B)x2+ (B−2C)x+C.
If this expression equals 1 for all x, then Cmust be 1 and the xand x2terms must cancel. Since C= 1 and
B−2C= 0 we have B= 2, and since A−B= 0 we also have A= 2. Therefore y= 2 + 1
x+1
x2is the only
solution of x4dy
dx +x2y= 1 of the desired form.
In fact, every solution of the equation has the form
y= 2 + 1
x+1
x2+Ce 1
x
for some constant C. It is possible to show this by rewriting the equation as
dy
dx +y
x2=1
x4,
then multiplying both sides by e−1
xand integrating.
2. Separate variables to get dy
y−2= (2x−1)dx, and then integrate both sides:
Zdy
y−2=Z(2x−1)dx =⇒ln |y−2|=x2−x+C.
To finish the problem we need to find C, get rid of the absolute values, and solve for y, not necessarily in
that order.
Method 1: plug in x= 1 and y= 5 to get
ln |5−2|= 12−1 + C=Cwhich implies C= ln 3.
Also, since y−2 is positive when x= 1 it must stay positive—it could not switch from positive to negative
without ln |y−2|becoming infinite. Therefore we can take the absolute values off, and we now have
ln(y−2) = x2−x+ ln 3.
Exponentiating both sides we get
eln(y−2) =ex2−x+ln 3 =ex2−xeln 3 which implies y−2 = 3ex2−x.
So we finally have y= 2 + 3ex2−x.
