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Differential Equations - Problem 2
To solve a differential equation with two initial conditions—that is, given d2y/dx2, find y—first take the integral of the expression with respect to x. The answer, dy/dx, is some expression + c. To solve for c, plug in the initial conditions y'(x0)=y'0. Then, plug in c to find a solution for dy/dx. By repeating this procedure (integrating and using the initial condition y(x0)=y0 to find c), solve for y.
Let’s try a slightly harder differential equation problem. Here I want to find the general solution of F’’(x) equals 2 over x cubed. Now remember, if I knew something about the second derivative of a function, I can get information about the first derivative by integrating.
So that’s what I’m going to do. We’ll get f’(x) by integrating 2 over x cubed with respect to x. Now I integrated this by first pulling the 2 outside, and rewriting this 1 over x cubed as x to the -3. I’m going to use the power rule on this.
The antiderivative of x then I get 3 x to the -2. Remember you add 1 to the exponent, -3 plus 1, -2. So this is 2 times x to the -2 and I have to divide by -2 as well. Then I add a constant. So this function becomes f’(x) equals, these cancel leaving –x to the -2, plus c. That’s f’(x), I want to find the f(x). So I have to integrate this again.
F(x) is the integral of –x to the -2 plus c, with respect to x. So first, I recognize that the above derivative of x to the -2 is x to the -1 over -1. So this is going to be minus because of this minus. X to the -1 over -1 and then when I antidifferentiate the plus c, the antiderivative of a constant is that constant times x. So it will be plus cx. And then I have to add another constant but don’t add plus c. You don’t want to be the same size as this. These might be different values, these two parameters. So change the name, call it d.
And so that means my final answer is f(x) equals, these negatives cancel, giving me x to the -1 plus cx plus d. This is my general solution to this differential equation. It's get two parameters that’s generally going to happen when you are solving a second order differential equation. That’s one of the reasons why we need two initial conditions to find a particular solution. And that’s what we are going to do next, so let’s take a look at part b.
Find the particular solution that satisfies f’(1) equals 3 and f(1) equals 0. So let’s recall two things; first, the f’(x) was –x to the -2 plus c. And second, that the final function was f(x) equals x to the -1 plus cx plus d.
You want to use both of these because you have an initial condition about f’ and f. Take a look at this equation. It only has one parameter, so this equation will allow me to solve for c. Let me use this initial condition when x is 1, f’ has to be 3. So I’ll write 3 equals -1 to the -2 plus c.
1 to the negative 2 is just 1 so this is minus 1. 3 equals minus 1 plus c. So add one to both sides and get c equals 4. So that’s one of my constants found. Look at this equation now I know that c is 4. I can use this second initial condition to solve for d. So when x is 1 f(x) is going to be 0. So this equation gives me 0 equals 1 to the -1 plus c times 1, now c is 4.
4 times 1 plus d. 1 to the -1 is 1. So this is 1 plus 4 , 5 . This is the same as 0 equals 5 plus d. So d equals -5. That means my particular solution comes from here, is f(x) equals x to the -1 plus cx. Remember c is 4, 4x, plus d, d was -5 so minus 5. This is my particular solution to my second order differential equation with two initial conditions.