1.5-6:
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.

1.5-18:
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).

1.5-22:
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:
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.