ENGR 213 Lecture Notes - Lecture 2: Elementary Function, Antiderivative

• School: Concordia University
• Department: Engineering
• Course: ENGR 213
• Professor: Georgios Vatistas

Document Summary

For first-order differential equations with only one constant p(y) dy dx p(y)dy = g(x) g(x)dx (1 + x)dy " ydx = 0 dy. Y ln y = ln 1 + x + c1. Y = 1 + x ec1 y = e (1 +c1 ln 1+x +c1 x) ln 1+x ec1 e (laws of exponents) y = c(1 + x) 2. 2 separable equations1separable equationmethod of solution cos x(e "2y dy: =dx (integration by parts) y = 0 when x = 0 implies c = 4. Solve the initial-value problem e sin 2x, y(0) = y. 2y e "y ey e cos x y gives dy = sin 2x cos x dx. )dy = e = "y sin xdx ye + "y. The solution of the initial-value problem is e +y. 4 " 2 cos x ye + "y e = "y. "x2 e x dy dt dt dt = 2 g(x) = e "x2 is continuous on the interval.