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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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