Consider the differential equation dy dx = x(y – 1).

Question

Consider the differential equation dy dx = x(y – 1). a) On the axes provided, sketch a slope field for the given differential equation at the 8 given points. b) Let y = f(x) be the particular solution to the given differential equation with the initial condition f(1) = 2. Write an equation for the line tangent to the graph of y = f(x) at x = 1. Use your equation to approximate f(1.5). c) Find the particular solution y = f(x) to the given differential equation with the initial condition f(1) = 2.

Answer 1

For part c)., here is the procedure I use to come up with final answer.

have ln|y-1|=(x²/2)+C,,,,,,the absolute value sign needed because the 1/(y-1) could have y<1 or y>1

since from initial condition know y>1, can remove absolute value sign and solve for C, ln(y-1)=(x²/2)-1/2

But say initial condition was (0,-2), then must rewrite ln(1-y)=(x²/2)+C and solve here C=ln3

Answer 2

Hey Izzy,

I probably can't answer this whole thing here, but I can help to get you started.

Slope fields are a bit of a pain to do by hand, but they aren't too hard.

All you need to do is plug in 8 different points into your equation for dy/dx.

For example, if we use the point (0,0) we have:

dy/dx = 0(0-1) = 0.

What this means is at the point (0,0), our solution curve is going to have a slope of 0. So you go to (0,0), and then graph a small line segment there that has a slope of 0 (horizontal).

From there, just repeat the process using 7 more points of your choosing.

For part b, you're looking for the equation for a tangent line. Remember, to write the equation for any line, we only need a slope and a point.

They've already told us that f(1) = 2, so we have our point (1,2). So all we need to know is what the slope of the tangent line would be at x = 1. This is where dy/dx comes in. We can plug (1,2) into our differential equation, and the output will be the slope of the tangent line. Now we can plug into point-slope form to find the equation of our line.

For part c, you can use your equation you came up with in part b. Just plug in 1.5 for x, and see what you get out for y.

Finally, part d is where we can actually solve a differential equation.

dy/dx = x(y-1) is separable, so we rearrange it like this:

(1/y-1)dy = x dx

Now you can integrate both sides, and then you can use your given initial condition to figure out what your constant of integration (C) should be.

I hope that helps!