# Homework Help: Homogeneous Differential equations

1. Nov 22, 2009

### Dell

in the following question i am asked to whow that Y1 and Y2 are basic solutions to the homogeneous equations

for 4.1)

y=ex=y' =y''

xy'' - (x+1)y' + y = x2
xex - (x+1)ex + ex = 0

none of these seem to work, what am i doing wrong?

2. Nov 22, 2009

### Dell

are these equations even homogeneous?

3. Nov 22, 2009

### Staff: Mentor

No, the given equations aren't homogeneous, but the equations formed by dropping the terms from the right sides (replacing the right sides by 0) are homogeneous.
For 4.1, the homogeneous equation is xy'' - (x + 1)y' + y = 0. It's easy to show that y_1 and y_2 are solutions to this homogeneous equation. By "basic" I think it means that y_1 and y_2 are linearly independent (easy to show) and span the space of solutions to the homogeneous equation, which is of dimension 2.

Same approach for 4.2.

4. Nov 22, 2009

### Dell

so what are they actually asking me to do? to show that by using y=e^x i can make the equation a homogeneous one??

5. Nov 22, 2009

### Staff: Mentor

No. Show that y = e^x and y = x + 1 are solutions to xy'' - (x+1)y' + y = 0. Note that these functions are NOT solutions to the nonhomogeneous equation, xy'' - (x+1)y' + y = x^2. The second problem is almost exactly the same.

Don't overthink this. You are not being asked to find solutions - just verify that the given functions are solutions to the homogeneous equations.

6. Nov 22, 2009

### Dell

but i dont understand why they would give me the nonhomogeneous equation

also am i supposed to use y1 and y2 seperately or at the same time, ie use y1 once prove that it is a solution and then use y2. or am i meant to use them together somehow

7. Nov 22, 2009

### Staff: Mentor

I guess because they are assuming you know the difference between a nonhomogeneous equation and the associated homogeneous equation.
Separately. However, if y1 and y2 are a set of basic solutions, which you need to show, then any linear combination of them is also a solution. IOW, if y1 and y2 are basic solutions to your 2nd order homogeneous DE, then y = c1y1 + c2y2 is a solution, and in fact represents all of them.