Originally Posted by

**harry** (a) Find by inspection particular solutions of the two nonhomogeneous equations

y ′′ + 2y = 4 and y′′ + 2y = 6x "Find by inspection"? Does this mean "take an educated guess at the answer"? If so, that shouldn't be too hard.

y'' + 2y = 4 ... Well, I see that this non-homogeneous equation involves only y and y'', and since the f(x) equation on the right is 4, I think we can safely assume the particular solution will just be a constant. Why? Because if it's a constant, then a constant times 2 is a constant (2y), and the second derivative of a constant is 0. So the question remains, then, what constant times 2 equals 4?

...

**y_p(x) = 2**

y'' + 2y = 6x ... Now f(x) is a linear function of x: f(x) = 6x. It seems pretty clear that y_p(x) will also be a linear funtion of x, such as y_p(x) = ax. Luckily, the second derivative of a linear funtion of x is 0, so now all we need is some value of a such that 2(ax) = 6x. You should find that a = 3. Therefore, **y_p(x) = 3x**

(b) find a particular solution of the differential equation

y′′ + 2y = 6x + 4.

Well, we have the particular solution to y'' + 2y = 4 and y'' + 2y = 6x. I'm sure you're teacher taught your class that the rule of superposition applies to linear, second order differential equations. This means that to find the particular solution of y'' + 2y = 6x + 4 we only need to add the particular solutions of y'' + 2y = 6x and y'' + 2y = 4. So the particular solution to this is **y_p(x) = 3x + 2**