Calculus question (differential equation)

Question

so I have a question about a differential equation..

So here's the differential equation: dy/dx = - 2x/y

The question wants me to find the particular solution y = f(x) to the differential equation with the initial condition f(1) = -1, and the correct answer is y = - sqrt( -2x +3).

I was wondering why there is a negative sign in front of the whole square root instead of just being positive.

Answer 1

Calculus question (differential equation)

so I have a question about a differential equation..

So here's the differential equation: dy/dx = - 2x/y

The question wants me to find the particular solution y = f(x) to the differential equation with the initial condition f(1) = -1, and the correct answer is y = - sqrt( -2x +3).

I was wondering why there is a negative sign in front of the whole square root instead of just being positive.

Hello Ashy,

Since, the differential dy/dx = -2x/y. Separate the variables and integrate.

dy/dx = -2x/y.

y*dy = -2x*dx

∫y*dy = -2 ∫x*dx

1/2y² = - x² + C

Now substitute x = 1 and f(1) = -1 to solve for C:

1/2(-1)² = - (1)² + C

1/2 + 1 = C

C = 3/2

The particular solution is simplified as follows:

1/2y² = - x² + C

1/2y² = - x² + 3/2

y² = 2( -x² + 3/2)

y² = -2x² + 3

y = +/- √(-2x² + 3)

The reason why the negative square root function is chosen over the positive, is based on the slope at the given point (1,-1). Note that dy/dx = - 2x/y represents the slope at any point (x,y) on the curve.

Notice that when plugging in x = 1 and y = -1 into the differential equation: dy/dx = - 2x/y

dy/dx = - 2x/y

dy/dx = - 2(1)/(-1)

dy/dx = + 2

This statement claims the slope of the tangent line at the point (1,-1) is a positive value of 2.

Recall that the graph y = +/- √(-2x² + 3), looks like a sideways parabola starting at 3 on the x-axis and opening to the left. There is an upper half and a lower half to this parabola. The reality is that the function forms an ellipse, with one half above the x-axis and the other half below.

Remember thopugh, you cannot have a function that violates the vertical line test, so we need to discard either the + or the - square root from the answer to keep it as a working function.

Since, the point (1,-1) is in the 4^th quadrant, this would infer that they only want the bottom half of the parabola/ellipse to write the function. If we used the + or upper portion, then the point (1,-1) is not in the range.

That is why they chose the - square root.

Any questions???