Answer:
The interval on which the solution is defined depends on the domain of the exponential function. Since e^((1/2)x^2 + ln(2)) is defined for all real numbers, the solution is defined on the interval (-∞, +∞), meaning the solution is valid for all x values.
Explanation:
o solve the differential equation dy/dx = xy with the initial condition y(0) = 2, we can separate the variables and integrate both sides.
Starting with the given differential equation:
dy/dx = xy
We can rearrange the equation to isolate the variables:
dy/y = x dx
Now, let's integrate both sides with respect to their respective variables:
∫(dy/y) = ∫x dx
Integrating the left side gives us:
ln|y| = (1/2)x^2 + C1
Where C1 is the constant of integration.
Now, we can exponentiate both sides to eliminate the natural logarithm:
|y| = e^((1/2)x^2 + C1)
Since y can take positive or negative values, we can remove the absolute value sign:
y = ± e^((1/2)x^2 + C1)
Next, we consider the initial condition y(0) = 2. Substituting x = 0 and y = 2 into the solution equation, we get:
2 = ± e^(C1)
Here, we see that e^(C1) is positive since it represents the exponential of a real number. So, the ± sign can be removed, and we have:
2 = e^(C1)
Taking the natural logarithm of both sides:
ln(2) = C1
Now, we can rewrite the general solution with the determined constant:
y = ± e^((1/2)x^2 + ln(2))