differential equation.?

Question

How do you determine wheter the given differential equation has the indicated function as solution?

2y'-y = 0

a. y1(x)= e^(x/2)

b. y2(x)= X^2 + 2e^(x/2)

I can find the general solution for that, but how do you tell if those functions are solutions?

I know that the first one is just by solving it for the dengeral solution. but how do you show that the second one is or is not?

Answer 1

a) Y' = (1/2)e^(x/2)
2Y' - Y = 2*(1/2)e^(x/2) - e^(x/2) = e^(x/2) - e^(x/2) = 0
So a) is a solution

b) Y'=2x + 2*(1/2)e^(x/2) = 2x + e^(x/2)
2Y' - Y = 2(2x + e^(x/2))-(X^2 + 2e^(x/2) = 4x - x^2
Therefor b) is not a solution

Answer 2

Compute 2y'_1-y_1 and 2y'_2-y_2 and simplify the results. If the result is identically the zero function, the function is a solution to the equation.

In your case, you have a homogeneous linear differential equation with constant coefficients so its general solution is determined by its characteristic equation, which in this case is 2k-1=0 ->k=1/2. Thus, $y=c e^(x/2)$ is a solution for all values of the constant c.

For you, this means that (a) is a solution. (b) is not a solution. To see that it is not a solution, observe that 2(x^2)''-x^2=8-x^2 \neq 0.

Good luck! :)

Answer 3

The best way is just to sub it in. So for a, you see that y = e^(x/2), and it's easy to calculate y' = (1/2)e^(x/2). So 2y' - y = 2(1/2)e^(x/2) - e^(x/2) = e^(x/2) - e^(x/2) = 0, so it is a solution. You can do the same for b to see if it is a solution.