The graph of a member of a family of solutions of a secondor

The graph of a member of a family of solutions of a second-order differential equation d^2y/dx^2 = f(x, y, y\') is given. Match the solution curve with at least one pair of the following initial conditions. Select one or more: y(2)=0, y\'(2) = -15 y(-1)=0, y\'(-1) = -6 y(0)=-6, y\'(0) = -5 y(2)=0, y\'(2)=15 y(-1)=0, y\'(-1) = 6 O y(0)= - 6, y\'(0) = 5

Solution

From the graph of the solution curve y = f(x), we can see that the curve intersects the x-axis at x = 2, x = – 1 and x = – 3. In other words, the curve passes through the points (2, 0), (– 1, 0) and (– 3, 0)

Therefore, y(2) = 0, y(– 1) = 0, and y(– 3) = 0

Also, the curve intersects the y-axis at y = – 6. In other words, the curve passes through the point (0, – 6)

Therefore, y(0) = – 6   

The first derivative y’(x) represent the slope of the curve at the point (x, y)

We will check the slopes of the curve at the points (0, – 6), (2, 0), (– 1, 0) i.e. at x = 0, x = 2, and x = – 1

If we draw slope at the point (0, – 6), the slope will create as angle measured counterclockwise from the positive x-axis which will be an obtuse angle. Therefore, slope at (0, – 6) = tan < 0

So, the best choice from the given options, is y’(0) = – 5

If we draw slope at the point (2, 0), the slope will create as angle measured counterclockwise from the positive x-axis which will be an acute angle. Therefore, slope at (2, 0) = tan > 0

So, the best choice from the given options, is y’(2) = 15

If we draw slope at the point (– 1, 0), the slope will create as angle measured counterclockwise from the positive x-axis which will be an obtuse angle. Therefore, slope at (– 1, 0) = tan < 0

So, the best choice from the given options, is y’(– 1) = – 6

Therefore, the selected initial conditions are:

d. y(2) = 0, y’(2) = 15

b. y(– 1) = 0, y’(– 1) = – 6

c. y(0) = – 6, y’(0) = – 5

(Answer)