Lecture12 - Math TAMU
... To find the specific solution that satisfies the initial conditions, we have to plug x = 1 into y(x) and y 0 (x). y(1) = (c1 + c2 )e2 = 1, y 0 (x) = (c2 + 2c1 + 2c2 x)e2x , y 0 (1) = (c2 + 2c1 + 2c2 )e2 = (2c1 + 3c2 )e2 = 1. So, we have system ...
... To find the specific solution that satisfies the initial conditions, we have to plug x = 1 into y(x) and y 0 (x). y(1) = (c1 + c2 )e2 = 1, y 0 (x) = (c2 + 2c1 + 2c2 x)e2x , y 0 (1) = (c2 + 2c1 + 2c2 )e2 = (2c1 + 3c2 )e2 = 1. So, we have system ...
