Differential Equations - Problem 3
Let’s take a look at another example of a differential equation. I’m asked to find the general solution of f’(x) equals 6x times the square root of x² plus 1. So remember if you have the information about the derivative, you can get the information about the function itself by integrating. So f(x) would be the integral of 6x times the square root of x² plus 1 dx.
This integral requires substitution. I have a radical function here, I should substitute for what’s inside the radical, the x² plus 1. So let me do that w equals x² plus 1, then dw will equal 2xdx. I have a 6xdx, that’s exactly 3 times what I need. So that gives me the square root of x² plus 1 is the square of w. And 6x dx is 3dw. I could pull the 3 out. And I can rewrite the square root as w to the ½. The antiderivative of the w to the ½ is w to the 3/2 divided by 3/2. Which is the same as w to the 3/2 times 2/3. I can’t forget this 3, plus a constant. So this simplifies to 2 times w to the 3/2 plus c. And now I re-substitute.
W was x² plus 1, so this becomes 2 times x² plus 1 to the 3/2 plus c. This is my f(x), my general solution to this differential equation. Now in part b, I’m asked to find a particular solution, the differential equation that satisfies f(0) equals 5. So let’s recall f(x) was 2 times x² plus 1 to the 3/2 plus c. I can find the value of c by using this initial condition. When I plug 0 in for x, I I should get 5. So 5 equals 2 times 0² plus 1 to the 3/2 plus c.
0² plus 1 is just 1, 1 to the 3/2 is just 1, and so this becomes 5 equals 2 plus c. Looks like c equals 3. And s my particular solution will be f(x) equals 2 times x² plus 1 to the 3/2 plus 3. This is the particular solution that says satisfies my differential equation and this initial condition.