By Paul Dawkins
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Additional info for Differential Equations
1 æp ö æp ö æp ö æp ö æp ö 3 2 = y ç ÷ = - cos ç ÷ cos ç ÷ - sin ç ÷ + c cos ç ÷ 2 è4ø è4ø è2ø è4ø è4ø 3 2=- 2 2 +c 2 2 c=7 The solution is then. 1 y ( x ) = - cos ( x ) cos ( 2 x ) - sin ( x ) + 7 cos ( x ) 2 Below is a plot of the solution. aspx Differential Equations Example 4 Find the solution to the following IVP. y (1) = t y¢ + 2 y = t 2 - t + 1 1 2 Solution First, divide through by the t to get the differential equation into the correct form. y¢ + 2 1 y = t -1+ t t Now let’s get the integrating factor, m ( t ) .
For this example the function that we need is Y ( x, y ) = y 2 + ( x 2 + 1) y - 3x3 Do not worry at this point about where this function came from and how we found it. Finding the function, Y(x,y), that is needed for any particular differential equation is where the vast majority of the work for these problems lies. As stated earlier however, the point of this example is to show you why the solution process works rather than showing you the actual solution process. We will see how to find this function in the next example, so at this point do not worry about how to find it, simply accept that it can be found and that we’ve done that for this particular differential equation.
Aspx Differential Equations ( 2 y - 4 ) dy = ( 3x 2 + 4 x - 4 ) dx ò ( 2 y - 4 ) dy = ò ( 3x 2 + 4 x - 4 ) dx y 2 - 4 y = x3 + 2 x 2 - 4 x + c We now have our implicit solution, so as with the first example let’s apply the initial condition at this point to determine the value of c. ( 3) 2 - 4 ( 3) = (1) + 2 (1) - 4 (1) + c 3 2 c = -2 The implicit solution is then y 2 - 4 y = x3 + 2 x 2 - 4 x - 2 We now need to find the explicit solution. This is actually easier than it might look and you already know how to do it.
Differential Equations by Paul Dawkins
Categories: Differential Equations