Updated 2013 1.5-6: Find the solution of xy'+5y=7x^2, y(2)=5 ====== 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. MAPLE ANSWER CHECK # xy'+5y=7x^2, y(2)=5 de:=x*diff(y(x),x)+5*y(x)=7*x^2; ic:= y(2)=5; dsolve([de,ic],y(x)); # Answer y(x) = (x^7+32)/x^5 1.5-10: Find the general solution of 2xy'-3y = 9x^3 ====== 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: Find the general solution of xy'=2y+x^3 cos(x) ====== 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). MAPLE ANSWER CHECK # xy'=2y+x^3 cos(x) de:=x*diff(y(x),x)=2*y(x)+x^3*cos(x); dsolve(de,y(x)); # Answer y(x) = x^2*sin(x)+x^2*_C1 # We write it as y = x^2 sin(x) + c x^2 1.5-20: Find the solution of y'=1+x+y+xy, y(0)=0 ====== 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. MAPLE ANSWER CHECK # y'=1+x+y+xy, y(0)=0 de:=diff(y(x),x)=1+x+y(x)+x*y(x); ic:= y(0)=0; dsolve([de,ic],y(x)); # Answer y(x) = -1+exp((1/2)*x*(x+2)) 1.5-22: Find the solution of y'=2xy+3x^2 exp(x^2), y(0)=5 ====== 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. Checkpoint: y=(x^3+c)exp(x^2). Use y(0)=5 to obtain c=5. MAPLE ANSWER CHECK # y'=2xy+3x^2 exp(x^2), y(0)=5 de:=diff(y(x),x)=2*x*y(x)+3*x^2*exp(x^2); ic:= y(0)=5; dsolve([de,ic],y(x)); # Answer y(x) = (x^3+5)*exp(x^2) 1.5-34: Use units of millions of cubic feet and days. ====== Reservoir volume 8000, initial pollution concentration of 0.25 percent, daily inflow 500 with pollution concentration 0.05 percent, and outflow 500. The fluid is well-mixed. Find the number of days required to reduce the concentration in the reservoir to 0.10 percent. 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 =(initial concentration)(volume)=(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_i)(c_i)=1/4. Then solve the initial value problem: x'=1/4-x/16, x(0)=20, using units of millions of cubic feet and days. The answer is x(t)=4+16 exp(-t/16). This is an easy-to-solve first order linear DE with constant coefficients. The answer is x(t) = x_h + x_p, where x_h=homogeneous solution = constant/(integrating factor) and x_p is the equilibrium solution. Model Derivation Law: x'=input rate - output rate. Definition: concentration == amt/volume. Use of percentages 0.25% concentration means 0.25/100 concentration The book's answer t = 16 ln 4 = 22.2 days is correct.