But, however, we don't have these remaining terms here. We only have, in this case, for the first order equation, the first term. So the solution then is equal to c1 e to the r1x But R 1, we already know here, so therefore the solution is C 1, e to the minus m x, is the solution to that equation, where C is some constant that we evaluate from the boundary conditions. The next special case are special order differential equations, which have this form Y double prime plus ay prime plus b is equal to 0, and in this case, we can solve this by the same general procedure that we just did, but the specific equations, the solutions to the second order, are given here, and I'll do an example on that. So the example here which of these is a solution to differential equation d two y by dx squared plus two dy by dx equals zero. Given the boundary conditions y of zero equals zero and y prime of zero, in other words dy by dx evaluated at zero is one. Which of these possible solutions Is the correct one. Well, first we compare our basic differential equation with the general statement of the linear homogeneous equation, which is this. And, in this case, comparing these, we see we only have the last three terms. Up here in this equation. And comparing this with this, we see that the coefficients are B2 is equal to one, B1 is equal to two, and B0 is equal to zero. So we follow the same steps as in the previous example. Next we go to our general statement of the characteristic equation, which is this. Realizing that we only have the last three terms appearing here because now this is a second order differential equation. So that equation becomes this, b2r squared plus b1r plus b0. Or r plus 2 is equal to 0. In other words, r equals -2. So, the two roots are r 1 equals 0 and r 2 equals -2. Now we go to our general solution, C 1, e to the r x, et cetera. But here, again, we recognize that we only have the first two terms in this equation will appear in the solution because it’s a second order differential equation. So the solution, then, is C1e to the R1x plus C2e to the R2x. And substituting these values in, we get this equation, which I can write like this. So this is our general solution. Next, however, we can evaluate those constants by applying the boundary conditions. So the first boundary condition i'll apply is this one, dy by dx evaluated at zero is equal to one. So, I differentiate this expression with respect to x to get minus two c two e to the minus 2x And then I substitute x=0 to evaluate that at 0. And I get that the value is equal to one. So therefore, solving that equation, the constant C2 is equal to minus one half. Next, I apply the first boundary condition, that y(0) = 0. So substituting x=0 into this equation, I get Y of zero is equal to C1 plus C2, from which it follows And evaluating that at 0, I have y(0) = 0, from which it follows that C1 is equal to -C2, but C2, we already know, is -1/2. Therefore, C1 = 1/2. So putting those values into the general solution here, we find that our solution is one-half minus one-half e to the minus 2x, which I can write a little more neatly as 1 minus e to the minus 2x over 2, and so the answer is b. And this completes our discussion of the solution of linear homogeneous differential equations.