I cant figure out how to do this type of problem!
Consider the first order differential equation y′+(x/(x^2−4))y=(e^x)/(x−9)
For each of the initial conditions below, determine the largest interval a<x<b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.
y(-6)=1.7
y(13)=-4
y(0)=0
y(-.5)=-4
y(4.5)=-.5
To determine the largest interval on which the existence and uniqueness theorem guarantees a unique solution for the given differential equation, we need to consider two factors:
1. Continuity: The function (x^2 - 4) in the denominator of the coefficient of y should not be equal to zero within the interval. If it is zero at any point, then the solution may not be unique.
2. Lipschitz condition: The function (x/(x^2 - 4)) in front of y should be Lipschitz continuous, meaning it should satisfy certain conditions that ensure the uniqueness of the solution.
To find the intervals satisfying these conditions, we first need to find the x-values where the denominator (x^2 - 4) is equal to zero. So, let's solve this equation:
x^2 - 4 = 0
Using the difference of squares, we have:
(x - 2)(x + 2) = 0
From this, we find that x = 2 and x = -2 are the points where the denominator is equal to zero.
Now, let's examine each initial condition one by one.
1. y(-6) = 1.7:
Here, the function is continuous throughout the entire real number line, and there are no points where the denominator becomes zero. Thus, there are no restrictions on the interval for this initial condition.
2. y(13) = -4:
Similar to the previous case, the function is continuous everywhere, and there are no points where the denominator is zero. So, no restrictions on the interval exist.
3. y(0) = 0:
In this case, notice that x = -2 is a point where the denominator is zero. Therefore, we must exclude this point from the interval. Hence, the largest interval is a < x < b, where a < -2 and b > -2.
4. y(-0.5) = -4:
In this case, x = -2 is again a point where the denominator is zero. Thus, we should exclude this point as well. So, the largest interval is a < x < b, where a < -2 and b > -2.
5. y(4.5) = -0.5:
In this final case, there are no points where the denominator becomes zero, and the function is continuous throughout the entire real number line. So, no restrictions on the interval exist.
To summarize the largest intervals for each initial condition:
1. y(-6) = 1.7: No restrictions, any interval.
2. y(13) = -4: No restrictions, any interval.
3. y(0) = 0: Interval is a < x < b, where a < -2 and b > -2.
4. y(-0.5) = -4: Interval is a < x < b, where a < -2 and b > -2.
5. y(4.5) = -0.5: No restrictions, any interval.