The slope field for a differential equation is shown in the figure. Determine the general solution of this equation.

The slope field has positive slopes in quadrants 2 and 4 and negative slopes in quadrants 1 and 4. It looks like a circle and at (0,0) it appears to have a slope of 1.

y=Cx2
x=Cy2
y2 – x2 = C2
x2 + y2 = C2

Well, it seems like the slope field is giving us a hint that the differential equation has something to do with circles. Let's clown around with some options:

How about y = Cx^2? This equation for a parabola doesn't quite match up with the slope field, so let's try something else.

Maybe x = Cy^2? This equation for a sideways parabola is also not in line with the slope field's circular shape.

Let's put our clown noses on and try one last option: y^2 - x^2 = C^2. This equation for a hyperbola seems to fit the bill! The positive slopes in quadrants 2 and 4 and negative slopes in quadrants 1 and 3 match up with the slope field's behavior.

But hold on, there's one more possibility: x^2 + y^2 = C^2. This equation for a circle seems like the perfect fit! The circular shape of the slope field and the slope of 1 at (0,0) make this the most likely solution.

So, my dear friend, it appears that the general solution for this differential equation is x^2 + y^2 = C^2. Now go and embrace your inner clown while solving more math problems!

To find the general solution of a differential equation given the slope field, you can integrate the slope field equation. In this case, the slope field indicates that the differential equation is separable, meaning you can separate the variables and integrate each side separately.

Let's start by separating the variables. We have:

dy/dx = y/x

To separate the variables, we can multiply both sides by dx and divide both sides by y:

(dy/y) = (dx/x)

Now we can integrate each side separately. Integrating the left side:

∫(dy/y) = ∫(dx/x)

This gives us:

ln|y| = ln|x| + C₁

Now we can solve for y by taking the exponential of both sides:

|y| = e^(ln|x| + C₁)

Using the properties of logarithms, we can simplify the right side:

|y| = e^(ln|x|) * e^(C₁)

Since e^(C₁) is just a constant, we can combine it with another constant. Let's call this new constant C:

|y| = C * |x|

Now we can consider the absolute values separately. Let's consider the case when y is positive:

y = Cx

And when y is negative:

-y = Cx

Now we have two separate solutions:

1. y = Cx, where C is any constant
2. -y = Cx, where C is any constant

Combining these two solutions, we get the general solution:

y = Cx or y = -Cx

Therefore, the general solution of this differential equation is y = Cx or y = -Cx.

If the slopes look like a circle, then it's a circle.

Note that for x^2+y^2 = 1
dy/dx = -x/y

bamboozled

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