INSTRUCTOR: Solve the initial value problem. The second derivative of y minus 3 times the first derivative of y, minus 10 times y, equals 0, with these conditions-- y of 0 equals 0, and y prime of 0 equals 7. Let's begin by considering the characteristic equation for this problem, this initial value problem, differential equation. This would be lambda squared minus 3 lambda minus 10 equals 0.

Now, because this factors as, open parentheses, lambda minus 5, close parentheses, open parentheses, lambda plus 2, close parentheses, equals 0, see that lambda is equal to negative 2, or 5. Because we have two distinct solutions, this means our general solution is y of x equal to c subscript of 1, times e to the negative 2x and exponent, plus c sub 2, followed by e to the 5x end exponent.

Now I'm going to go ahead and find the first derivative of this because we will need this in just a second. So y prime of x is equal to negative 2 times c sub 1, times e to the negative 2x, and exponent plus 5 times c sub 2, times e to the 5x.

Now for y of x, we know that y of 0 equals 0, which gives us this equation, 0 equals c sub 1 plus c sub 2. And for y prime of 0 equal to 7, evaluating that we see 7 equals negative 2 times c sub 1, plus 5 times c sub 2.

Combining these together, we have the system of equations c sub 1 plus c sub 2 equals 0, and negative 2 times c sub 1 plus 5 times c sub 2 equals 7. And I would use some elimination here. I'd multiply the first equation by 2, and then add them together. And this leads us to see that c sub 1 equals negative 1, and c sub 2 equals positive 1.

So we had the general solution, form y of x, but now we can say what the specific solution is for those conditions. Y of x equals negative e to the negative 2x and exponent, plus e to the 5x end exponent. Again, that's because we know c1 and c2, so we can replace those values.

And that is the solution to that initial value problem.