To determine the values of r for which the function y = erx satisfies the differential equation 2yy’ = y, we first need to differentiate the function with respect to x.
The function y = erx is an exponential function, where e is the base of the natural logarithm. The derivative of y with respect to x is:
y’ = r imes e^{rx}
Now, substituting y and y’ into the differential equation:
2yy’ = 2(erx)(r e^{rx}) = 2r e^{2rx}
On the right side of the equation, we have:
y = erx
Therefore, we equate both sides:
2r e^{2rx} = e^{rx}
To solve for r, we can simplify this equation by dividing both sides by e^{rx} (assuming e^{rx}
eq 0):
2r e^{rx} = 1
Now we can isolate e^{rx}:
e^{rx} = rac{1}{2r}
For this equation to hold true, the term 2r must be non-zero:
r
eq 0
Furthermore, since e^{rx}
e 0 for any real r, there are no additional restrictions on r. Thus, the values of r that satisfy the original differential equation are all real numbers, except zero. In conclusion:
All real values of r except r = 0 satisfy the differential equation.