Calculus Volume 1

494

Chapter 4 | Applications of Derivatives

Figure 4.86 Some of the solution curves of the differential equation dy dx =6 x 2 are displayed. The function y =2 x 3 +3 satisfies the differential equation and the initial condition y (1) =5.

Example 4.53 Solving an Initial-Value Problem

Solve the initial-value problem

dy dx = sin

x , y (0) =5.

Solution First we need to solve the differential equation. If dy

dx = sin x , then y = ∫ sin( x ) dx =−cos x + C . Next we need to look for a solution y that satisfies the initial condition. The initial condition y (0) =5 means we need a constant C such that −cos x + C =5. Therefore, C = 5 + cos(0) = 6. The solution of the initial-value problem is y =−cos x +6.

dy dx =3

4.52

x −2 , y (1) =2.

Solve the initial value problem

Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car.

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