Solutions to ODEs

Solution to Equations in Algebra I In algebra, an equation is two expressions involving a single variable (usually x) that are set equal to each other; such as x 2 = 3x − 2. I If an equation such as x 2 = 3x − 2 and we seek a solution to this equation, then we mean to say ‘find all possible values of x that satisfy this equation, i.e. when you plug the value of x into the equation then the Left-Hand-Side (LHS) of the equation has the same output as the Right-Hand-Side (RHS) of the equation. I For instance, if x = 1 then the LHS will be 12 which simplifies to 1 and the RHS will be 3 · 1 − 2 which simplifies to 1 as well. Thus, the LHS and the RHS agree for this value of x. So, x = 1 is a solution. I Similarly, if x = 2 then LHS and RHS will both equal 4. So x = 2 is a solution. Definition Consider the ODE dy dx = f (x, y ). By a solution to the ODE, we mean a FUNCTION y = g (x) that satisfies the ODE. 0 I LHS is dy dx = g (x). I RHS is f (x, y ) = f (x, g (x)). I If LHS = RHS after simplifying, then y = g (x) is a solution to the ODE (FOR ALL x IN THE DOMAINS OF y AND dy dx )! Example x Consider the ODE dy dx = y + x. Show that y = −x − 1 + Ce is a solution to the ODE, where C is an arbitrary constant term. Solution: d x x x I LHS is dy dx = dx [−x − 1 + Ce ] = −1 − 0 + Ce = −1 + Ce . I RHS is y + x = (−x − 1 + Ce x ) + x = −1 + Ce x . I Since LHS = RHS, we have that y = −x − 1 + Ce x satisfies the ODE and, hence, is a solution to the ODE. I Notice that at no point did we assume a value for the arbitrary constant C , hence it must be true regardless of the value of C . I Thus, there are infinitely many solutions to the ODE as we can plug in whatever we want for C . Example 2 Consider the ODE ddxy2 + y = 0. Show that y = C1 cos(x) + C2 sin(x) is a solution to the ODE, where C1 and C2 are arbitrary constants. Solution: I LHS is complicated, so let’s first compute y dy dx d 2y dx 2 d 2y . dx 2 = C1 cos(x) + C2 sin(x) d = [C1 cos(x) + C2 sin(x)] = −C1 sin(x) + C2 cos(x) dx   d dy = = −C1 cos(x) − C2 sin(x) dx dx Example 2 Consider the ODE ddxy2 + y = 0. Show that y = C1 cos(x) + C2 sin(x) is a solution to the ODE, where C1 and C2 are arbitrary constants. Solution: I Thus, LHS simplifies to... d 2y +y dx 2 = (−C1 cos(x) − C2 sin(x)) +(C1 cos(x) + C2 sin(x)) = 0. Example 2 Consider the ODE ddxy2 + y = 0. Show that y = C1 cos(x) + C2 sin(x) is a solution to the ODE, where C1 and C2 are arbitrary constants. Solution: I Note that RHS is also 0. Thus, y = C1 cos(x) + C2 sin(x) is a solution for all values of C1 and C2 . Solutions to an IVP I We now know what solutions to ODEs look like. I Recall that an nth order IVP has an ODE and n additional pieces of information about the dependent variable y . I In order to be a solution to an IVP, the function must satisfy all n + 1 of these equations. Example Show that y = cos(x) + sin(x) is a solution to the IVP: y 00 + y = 0, y (0) = 1, y 0 (0) = 1. Solution: I In the previous example, we showed that y = C1 cos(x) + C2 sin(x) is a solution to the ODE y 00 + y = 0 for any choice of C1 and C2 . Example Show that y = cos(x) + sin(x) is a solution to the IVP: y 00 + y = 0, y (0) = 1, y 0 (0) = 1. Solution: I Then, letting C1 = 1 and C2 = 1 we see that y = cos(x) + sin(x) is a solution to the ODE without having to show any more work (as we showed it in the previous problem). Example Show that y = cos(x) + sin(x) is a solution to the IVP: y 00 + y = 0, y (0) = 1, y 0 (0) = 1. Solution: I Now, we must show that it satisfies y (0) = 1 and y 0 (0) = 1. We will start with y (0) = 1. y (0) = cos(0) + sin(0) = 1 + 0 = 1 Example Show that y = cos(x) + sin(x) is a solution to the IVP: y 00 + y = 0, y (0) = 1, y 0 (0) = 1. Solution: I We next show y 0 (0) = 1. Notice that we first need to derive that y 0 (x) = − sin(x) + cos(x) before we can show this. y 0 (0) = − sin(0) + cos(0) = 0 + 1 = 1 Example Show that y = cos(x) + sin(x) is a solution to the IVP: y 00 + y = 0, y (0) = 1, y 0 (0) = 1. Solution: I Then, y = cos(x) + sin(x) satisfies all three equations in the IVP and, hence, is a solution to the IVP.