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). 1.5-20: ====== 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: ====== 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.