- (a) The integrating factor is $ e^{\int 2x \, dx} = e^{x^2} $. Multiplying both sides of the DE by this factor gives $ e^{x^2} \frac{dy}{dx} + 2x e^{x^2} y = x e^{x^2} $. The left side is the derivative of $ y e^{x^2} $, so $ \frac{d}{dx}(y e^{x^2}) = x e^{x^2} $. Integrating both sides yields $ y e^{x^2} = \frac{1}{2} e^{x^2} + C $, so the solution is $ y = \frac{1}{2} + C e^{-x^2} $.
- (b) Let $ z = \frac{dy}{dx} $. The two first-order DEs are:
(i) $ \frac{dz}{dx} + 2z + y = x $
(ii) $ \frac{dy}{dx} = z $
- (c)(i) The auxiliary equation is $ m^2 + 2m + 1 = 0 $, which factors as $ (m+1)^2 = 0 $. The repeated root is $ m = -1 $. Therefore, the general solution to the homogeneous equation is $ y_h = (C_1 + C_2 x) e^{-x} $.
- (c)(ii) For the nonhomogeneous DE $ \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + y = 2x + 4 $, assume a particular solution of the form $ y_p = Ax + B $. Then $ y_p' = A $ and $ y_p'' = 0 $. Substituting into the DE gives $ 0 + 2A + (Ax + B) = 2x + 4 $, so $ Ax + (2A + B) = 2x + 4 $. Equating coefficients: $ A = 2 $, $ 2A + B = 4 $ → $ 4 + B = 4 $ → $ B = 0 $. Thus, $ y_p = 2x $.
- (c)(iii) The general solution is the sum of the homogeneous solution and the particular solution: $ y = y_h + y_p = (C_1 + C_2 x) e^{-x} + 2x $.
Parent Tip: Review the logic above to help your child master the concept of separable differential equations worksheet.
