# S13 - 1.3 Initial Conditions Initial-Value Problems As we...

This preview shows pages 1–3. Sign up to view the full content.

1.3 Initial Conditions; Initial-Value Problems As we noted in the preceding section, we can obtain a particular solution of an n th order diﬀerential equation simply by assigning speciFc values to the n constants in the general solution. However, in typical applications of diﬀerential equations you will be asked to Fnd a solution of a given equation that satisFes certain preassigned conditions. Example 1. ±ind a solution of y ± =3 x 2 - 2 x that passes through the point (1 , 3). SOLUTION In this case, we can Fnd the general solution by integrating: y = ± ( 3 x 2 - 2 x ) dx = x 3 - x 2 + C. The general solution is y = x 3 - x 2 + C . To Fnd a solution that passes through the point (2 , 6), we set x = 2 and y = 6 in the general solution and solve for C : 6=2 3 - 2 2 + C =8 - 4+ C which implies C =2 . Thus, y = x 3 - x 2 + 2 is a solution of the diﬀerential equation that satisFes the given condition. In fact, it is the only solution that satisFes the condition since the general solution represented all solutions of the equation and the constant C was uniquely determined. ± Example 2. ±ind a solution of x 2 y ±± - 2 xy ± +2 y =4 x 3 which passes through the point (1 , 4) with slope 2. SOLUTION As shown in Example 4 in the preceding section, the general solution of the diﬀerential equation is y = C 1 x 2 + C 2 x x 3 . Setting x = 1 and y = 4 in the general solution yields the equation C 1 + C 2 + 2 = 4 which implies C 1 + C 2 . The second condition, slope 2 at x = 1, is a condition on y ± . We want y ± (1) = 2. We calculate y ± : y ± C 1 x + C 2 +6 x 2 , and then set x = 1 and y ± = 2. This yields the equation 2 C 1 + C 2 + 6 = 2 which implies 2 C 1 + C 2 = - 4 . Now we solve the two equations simultaneously: 11

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
C 1 + C 2 =2 2 C 1 + C 2 = - 4 We get: C 1 = - 6 ,C 2 = 8. A solution of the diﬀerential equation satisfying the two conditions is y ( x )= - 6 x 2 +8 x +2 x 3 . It will follow from our work in Chapter 3 that this is the only solution of the diﬀerential equation that satisFes the given conditions. ± INITIAL CONDITIONS Conditions such as those imposed on the solutions in Examples 1 and 2 are called initial conditions. This term originated with applications where processes are usually observed over time, starting with some initial state at time t =0 .
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 06/20/2008 for the course M 427K taught by Professor Fonken during the Spring '08 term at University of Texas.

### Page1 / 7

S13 - 1.3 Initial Conditions Initial-Value Problems As we...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online