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1.3 Initial Value Problems An initital value problem consists of the following information: 1. A differential equation involving an unknown function y. Instead of including y (0) , sometimes an initial value problem includes a different value of y, such as y (1) , or possibly the limiting value of y as x → −∞ or x → ∞. 2. An initial value for y, i.e. the value of y when x 0. Since y is a function, the value of y when x 0 is usually denoted y (0) . The idea of an initial value problem is that specifying the value of y (0) is usually enough information to determine a single solution to the differential equation. Usually, an initial value problem has only one solution. Here is the most common procedure for solving an initial value problem. Solving Initial-Value Problems 1. Find the general solution to the given differential equation, involving an arbitrary constant C. 2. Plug in the initial value to get an equation involving C, and then solve for C. EXAMPLE 1 Find the solution to the following initial value problem: y 0 −y 2 , SOLUTION y (0) 3. We previously found the general solution to this equation, namely y 1 , x+C where C is an arbitrary constant. Therefore, we need only find a value for C for which y (0) 3. Plugging in x 0 and y 3 gives the equation 3 1 . 0+C Solving for C gives C 1/3, so 1 . x + 1/3 We can simplify our answer by multiplying the numerator and denominator by 3: y y 3 3x + 1 EXAMPLE 2 Find the solution to the following initial value problem: y 0 2y, y (0) 5. INITIAL VALUE PROBLEMS SOLUTION 2 The given differential equation isn’t very different from the equation y 0 y. In that case, the general solution was y Ce x . How can we modify this solution to account for the extra 2? A few moments of thought reveals the answer: More generally, the solution to any equation of the form y 0 k y (where k is a constant) is y Ce kx . y Ce 2x So this is the general solution to the given equation. Plugging in x 0 and y 5 gives the equation 5 Ce 0 , so C 5, and the solution is y 5e 2x . EXERCISES 1–2 Solve the given initial value problem. 1. y 0 xe x , y (0) 3 2. y 0 3y, y (2) 4