###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Differential Equations - Problem 1

Norm Prokup
###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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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 take a look at a differential equation with initial condition. Solve the second derivative of y with respect to x, equals 6x plus 18, if y equals 4 and dy/dx equals 12 when x equals 1. This actually counts as two initial conditions. In general when you are solving the second order differentiation equation you’ll need two initial conditions.

Well let’s start with this differential equation. Second derivative of y with respect x equals 6x plus 18. In general we want to integrate it down, so from second derivative to first derivative. If this is the second derivative, to get the first derivative we integrate. So dy/dx will equals the integral of 6x plus 8 with respect to 18.

So that’s what we have to integrate. The interval of 6x is 3x², the interval of 18 is 18x plus c. I need to use one of my initial conditions hopefully to find the value of this constant. It’s this one dy/dx is 12 when x equals 1. So I plug in 12 for dy/dx when x equals 1. 3 times 1² plus 18 times 1 plus c. Here this adds up to 21, so this is just 12 equals 21 plus c. So c would have to equal -9. And what that gives me is a particular value for the dy/dx. Dy/dx I now know is 3x² plus 18x minus 9.

So I have to integrate again to find y. This will be the solution I’m looking for. So I want to find y subject to this condition. That’s another differential equation which I’ll start up here. So dy/dx equals 3x² plus 18x minus 9. I find y by integrating.

So this is going to be y equals the integral of this thing; 3x² plus 18x minus 9. So y equals the antiderivative of 3x² is 3 times 1/3 x cubed or x cubed, the antiderivatives of 18x is going to be 18 times ½ x² so 9x². The anti derivative of -9 is -9x. And I need to add another integration constant. Only this time I want to use a different letter because it’s probably going to have a different value in c. So I’m going to use plus d.

Now how do I find d? I have to use the second initial condition; y equals 4 when x equals 1. So let’s write that down. Y equals 4 when x equals 1. So I plug in 4, I plug in 1 for x, 9 times 1² minus 9 times 1 plus d. This is 9 minus 9, that’s just going to cancel. This is 1 which I can subtract and get d equals 3. That’s my value for d.

So my final answer is y equals x cubed plus 9x² minus 9x plus 3. This is the particular solution to my original second order differential equation with initial conditions.