Question

Solution of the differential equation \(x = 1 + xy\frac{dy}{dx} + \frac{(xy)^2}{2!}\left(\frac{dy}{dx}\right)^2 + \frac{(xy)^3}{3!}\left(\frac{dy}{dx}\right)^3 + ..........\) is

• A. \(y = e^{\log x} + C\)
• B. \(y = e^{(\log x)^2} + C\)
• C. \(y = \pm e^{\frac{(\log x)^2}{2}} + C\)
• D. \(xy = x + C y\)

Hint:

Recognize the exponential series: \(e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + ..........\).

Answer 1

The Correct Option is C

The problem involves solving a differential equation given in series form as: 

\(x = 1 + xy\frac{dy}{dx} + \frac{(xy)^2}{2!}\left(\frac{dy}{dx}\right)^2 + \frac{(xy)^3}{3!}\left(\frac{dy}{dx}\right)^3 + \ldots\)

We can observe that this series resembles the Taylor expansion of the exponential function \(e^u\), where:

\(e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \ldots\)

Comparing this with the given series, we find:

\(u = xy \left(\frac{dy}{dx}\right)\)

Therefore, the differential equation can be rewritten as the exponential equation:

\(e^{xy\frac{dy}{dx}} = x\)

Taking natural logarithms on both sides, we get:

\(xy \frac{dy}{dx} = \ln x\)

Separating variables, we have:

\(y \, dy = \frac{\ln x}{x} \, dx\)

Integrating both sides, the left side becomes:

\(\int y \, dy = \frac{y^2}{2} + C_1\)

For the right side, the integral of \(\frac{\ln x}{x}\) is a known result:

\(\int \frac{\ln x}{x} \, dx = \frac{(\ln x)^2}{2} + C_2\)

Equating the integrals gives:

\(\frac{y^2}{2} = \frac{(\ln x)^2}{2} + C\)

where \(C = C_2 - C_1\).

After simplifying and solving for \(y\), we get:

\(y^2 = (\ln x)^2 + 2C\)

This implies:

\(y = \pm \sqrt{(\ln x)^2 + 2C}\)

Renaming \(2C\) as y = \pm e^{\frac{(\ln x)^2}{2}} + C, which matches the correct answer.

Thus, the solution is:

\( y = \pm e^{\frac{(\log x)^2}{2}} + C \)