1.5-6:
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The standard form of the equation is y'+(5/x)y = 7x
Let W be the integrating factor W=exp(int(p(x),x)), simplify to W=x^5.
Replace y'+(5/x)y by (Wy)'/W, clear fractions, apply quadrature.
According to the book's answers, this gives y=x^2+c/x^5. 
Use y(2)=5 to obtain c=32.

1.5-10:
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The standard form of the equation 2xy'-3y = 9x^3 is y'-(1.5/x)y = 4.5x^2
or y' + py = q with p(x)=-1.5/x, q(x)= 4.5x^2. 

  Let W be the integrating factor W=exp(int(p(x),x)) and simplify to 
  W=x^(-1.5) using n ln(x) = ln(x^n) with n=-1.5 and exp(ln(u))=u 
  where u=x^(-1.5).

  Replace the standard form equation by (Wy)'/W = q. Cross-multiply and
  apply the method of quadrature. Then Wy=C+int(qW) reduces to the
  integration problem int(qW). 

  Hence you must find int(F(x)) for F(x)=4.5(x^2)(x^(-1.5)). This is a 
  power integral problem and the answer is (4.5/1.5) x^(1.5).

  Final answer: Wy=C+3x^(1.5) which simplifies to y = C x^(1.5) + 3 x^3.

Maple answer check: 
  de:=2*x*diff(y(x),x)-3*y(x) = 9*x^3;
  dsolve(de,y(x));

1.5-18:
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The standard form of the equation is y'+(-2/x)y = x^2 cos(x).
Let W be the integrating factor W=exp(int(p(x),x)), simplify to W=1/x^2.
Replace y'+(-2/x)y by (Wy)'/W, clear fractions, apply quadrature.
The book's answer is y=x^2(sin(x)+c).

1.5-20:
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The standard form of the equation is y'+(-x-1)y = x+1. Let W be the
integrating factor W=exp(int(p(x),x)). Simplify to W=exp(-x^2/2-x).
Replace y'+(-x-1)y by (Wy)'/W, clear fractions, apply quadrature. To
integrate RHS=(x+1)W=(x+1)exp(-x^2-x), use substitution u=-x^2-x, then
(x+1)W=exp(u)(-du) which integrates to -exp(u) or -exp(-x^2/2-x), which
is -W! After quadrature, Wy=-W+c, and then y=-1+c/W. Use y(0)=0 to
obtain c=1 and finally y = -1 + exp(x^2/2+x). The book's answer is wrong
in Edwards-Penney 2/E but corrected in 3/E. The solution manual in 2/E
also has an error.

1.5-22:
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The standard form of the equation is y'+(-2x)y = 3 x^2 exp(x^2).
Notation: exp(u) means e^u. For example, exp(-1) is 1/e.
Let W be the integrating factor W=exp(int(p(x),x)), simplify to W=exp(-x^2).
Replace y'+(-2x)y by (Wy)'/W, clear fractions, apply quadrature.
The book answers imply y=(x^3+c)exp(x^2).
Use y(0)=5 to obtain c=5.

1.5-34:
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The units should be taken as millions of cubic feet.

The textbook gives the initial value problem x'=r_i c_i - (r_0/v)x,
x(0)=x_0. 

The initial value is x_0 =(0.25/100)8000, the output rate is r_0=500,
and the tank volume is V=8000.

Please determine the value for the input concentration, constant c_i.
You should obtain r_ic_i=1/4.

Then solve the initial value problem.
The book's answer t =  16 ln 4 = 22.2 days is correct.