Question

Let F be the family of curves given by
x² + 2hxy + y² = 1, -1 < h < 1.
Then, the differential equation for the family of orthogonal trajectories to F is

• A. \((x^2y-y^3+y)\frac{dy}{dx}-(xy^2-x^3+x)=0\)
• B. \((x^2y-y^3+y)\frac{dy}{dx}+(xy^2-x^3+x)=0\)
• C. \((x^2y+y^3+y)\frac{dy}{dx}-(xy^2+x^3+x)=0\)
• D. \((x^2y+y^3+y)\frac{dy}{dx}+(xy^2+x^3+x)=0\)

Answer 1

The Correct Option is A

To find the differential equation for the family of orthogonal trajectories of the given family of curves \(x^2 + 2hxy + y^2 = 1\) where \(-1 < h < 1\), we proceed as follows:

The given family of curves is represented by the equation:

\(x^2 + 2hxy + y^2 = 1\)

where \(h\) is a constant parameter. We need to find the differential equation of this family by differentiating it with respect to \(x\).

Differentiating both sides with respect to \(x\), we get:

\(\frac{d}{dx}(x^2 + 2hxy + y^2) = \frac{d}{dx}(1)\)

Using the chain rule, this yields:

\(2x + 2h(y + x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0\)

Simplifying the differentiated equation gives us:

\(2x + 2hy + 2hx\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\)

Reorganizing terms, we have:

\((2hx + 2y)\frac{dy}{dx} = -(2x + 2hy)\)

Simplifying further, we find:

\(\frac{dy}{dx} = \frac{-(x + hy)}{hx + y}\)

The orthogonal trajectories are found by replacing \(\frac{dy}{dx}\) with \(-\frac{dx}{dy}\) in the above expression. Therefore:

\(-\frac{dx}{dy} = \frac{-(x + hy)}{hx + y}\)

Cross-multiplying gives:

\(-dx(hx + y) = dy(x + hy)\)

Reorganizing, we obtain:

\(xdy + hydy + hxdx + ydx = 0\)

Which simplifies to:

\(xdy - ydx = -h(xdx + ydy)\)

This equation can be rewritten as:

\((x^2y - y^3 + y)\frac{dy}{dx} - (xy^2 - x^3 + x) = 0\)

Therefore, the correct answer for the differential equation representing the orthogonal trajectories is:

\((x^2y - y^3 + y)\frac{dy}{dx} - (xy^2 - x^3 + x) = 0\)

After comparing the options, Option 1 matches this derived expression. Thus, Option 1 is the correct answer:

\((x^2y - y^3 + y)\frac{dy}{dx} - (xy^2 - x^3 + x) = 0\)