Integrating Factor Examples Shortcut Examples with 60-second solutions Long Example with full integration =================================================================== EXAMPLE. Solve y' + 2y = 6 SOLUTION. Theory says that y=y_h + y_p, and because the coefficients are constant, then y_p = an equilibrium solution. Therefore constant y = ------------------ + equilibrum solution integrating factor EQUILIBRIUM SOLUTION. Formally set y' to zero, getting y' + 2y = 6 original DE 0 + 2y = 6 set y' to zero because an equil sol is constant y = 3 Equilibrium solution found, this is y_p INTEGRATING FACTOR. Theory says it is W = exp(int(p(x),x)) for equation y'+p(x)y=q(x) W = exp(int(2,x)) W = exp(2x) GENERAL SOLUTION. Use symbol C for the constant in the homogeneous solution y_h, then constant y = ------------------ + equilibrum solution integrating factor C y = ------- + 3 or y = C exp(-2x) + 3 exp(2x) =================================================================== PROBLEM. Solve y' + 3y = 6 ANSWER: y = C exp(-3x) + 2 How long did it take? After practice, about 60 seconds. =================================================================== PROBLEM. Solve 2y' + 5y = 3 ANSWER: y = C exp(-5x/2) + 3/5 Divide by 2 first to get the form y'+py=q. =================================================================== PROBLEM. Solve 2y' + sin(x)y = 0 ANSWER: y = C exp(cos(x)/2) THEOREM. The solution of y'+py=0 is y=C/integrating factor. Divide by 2 first to get the form y'+py=0. Then find the integrating factor W for p=sin(x)/2. Then y_h = C/W. Confused by y_p? Select y_p=0 (it works) in the superposition formula y = y_h + y_p. More work for the same answer. =================================================================== PROBLEM. Solve xy' + 2y = 0 ANSWER: y = C exp(-2 ln|x|) or y=C/x^2 THEOREM. The solution of y'+py=0 is y=C/integrating factor. Divide by x first to get the form y'+py=q. Then find the integrating factor W for p = 2/x. Then y_h = C/W. =================================================================== PROBLEM. Solve xy' + 2y = x^2 ANSWER: y = C/x^2 THEOREM. For y'+py=q, y=y_h + y_p (superposition) THEOREM. The solution of y'+py=0 is y_h=C/W, W=integrating factor. THEOREM. A particular solution of y'+py=q is y_p=(1/W)int(qW,x). Divide by x first to get the form y'+py=q, p=2/x, q=x^2/x Then find the integrating factor W=x^2 for p=2/x. Then y_h = C/W = C/x^2 Integrate int(qW,x) where q=x^2/x (divided by x) and W=x^2. Answer: int(qW,x)=int(x^3,x)=x^4/4 y_p=(1/W) int(qW,x) = (1/x^2)(x^4/4) = x^2/4 y = y_h + y_p = C/x^2 + x^2/4 Answer check: xy' + 2y = x[(C/x^2)' + (x^2/4)'] + 2y of Original DE = -2xC/x^3 + 2x^2/4 + 2C/x^2 + 2x^2/4 = x^2