Self Quiz Solutions 28 - Introduction to Differential Equations | MATH 220, Study notes of Differential Equations

Download Self Quiz Solutions 28 - Introduction to Differential Equations | MATH 220 and more Study notes Differential Equations in PDF only on Docsity! MATH 220 Self-quiz 28 1. Find the general solution of the differential equation: y′ = 2y − sin x −2x + sin y Solution: We can rewrite the equation as dy dx = 2y − sin x −2x + sin y and then cross multiply.: (2y − sin x)dx = (−2x + sin y)dy (2y − sin x)dx + (2x− sin y)dy = 0 This is an exact equation. Indeed if M = 2y − sin x and N = 2x − sin y, then: My = 2 Nx = 2 Therefore, we find the potential as: ϕ(x, y) = ∫ (2y − sin x)dx = 2xy + cos x + f(y) Now, we need to figure out what f(y) is: ∂ ∂y (2xy + cos x + f(y)) = 2x− sin y 2x + f ′(y) = 2x− sin y f(y) = − ∫ sin ydy = cos y Therefore: ϕ(x, y) = 2xy + cos x + cos y and the general solution of the original equation can be given in the form: 2xy + cos x + cos y = C 2. Solve the initial value problem: y′′ + 4y′ + 3y = e−x y(0) = 1 y′(0) = 0 Solution: We begin by solving the corresponding homogeneous equation. Its auxillary equation is r2 + 4r + 3 = 0 with roots −3 and r− 1. Thus, the solution of the homogeneous is: yh = C1e−3x + C2e−x Now, we will seek a particular solution of the non-homogeneous equation. This should have the form yp = axe−x Its derivatives are: y′p = −axe−x + ae−x y′′p = axe−x − 2ae−x By plugging these derivatives back into the original equation, we get axe−x − 2ae−x − 4axe−x + 4ae−x + 3axe−x = 2ae−x By comparing coefficients, we can find a: 2ae−x = e−x =⇒ 2a = 1 =⇒ a = 1 2 2