Question

If \( x \log_{10} x \frac{dy}{dx} + y = 2 \log_{10} x \) and \( y(e) = 0 \), then \( y(e^2) = \)

• A. 0
• B. 1
• C. \( \frac{2}{3} \)
• D. \( \frac{3}{2} \)

Hint:

For linear differential equations, use the integrating factor \( e^{\int P(x) \, dx} \). Apply initial conditions to find the constant and evaluate at the desired point.

Answer 1

The Correct Option is D

To solve the given differential equation, we start by interpreting the equation:
\( x \log_{10} x \frac{dy}{dx} + y = 2 \log_{10} x \).

We can rewrite this in a separable form:
\((x \log_{10} x) \frac{dy}{dx} = 2 \log_{10} x - y\).

Separating variables, we get:
\(\frac{dy}{dx} = \frac{2 \log_{10} x - y}{x \log_{10} x}\).

We apply integration considering the initial condition \( y(e) = 0 \).
For solving we can employ the integrating factor method. The integrating factor \( \mu(x) \) for a linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \) is given by \( e^{\int P(x) dx} \).

In this case, comparing:
- \( P(x) = \frac{1}{x\log_{10} x} \)
- \( Q(x) = \frac{2 \log_{10} x}{x \log_{10} x} = \frac{2}{x} \).

The integrating factor is:
\(\mu(x) = e^{\int \frac{1}{x \log_{10} x} dx}\).

This integral simplifies to:
\( \int \frac{1}{x \log_{10} x} dx = \log_{10}(\log_{10} x) \).
Thus, the integrating factor becomes:
\(\mu(x) = e^{\log_{10}(\log_{10} x)} = \log_{10} x\).

Multiplying through the differential equation by the integrating factor yields:
\(\log_{10} x \frac{dy}{dx} + \frac{\log_{10} x}{x \log_{10} x} y = \frac{2 \log_{10} x}{x}\).
This equation simplifies to:
\(\frac{d}{dx} (y \log_{10} x) = \frac{2}{x} \log_{10} x\).

Integrating both sides w.r.t \( x \):
\( y \log_{10} x = \int \frac{2}{x} \log_{10} x \, dx \).

Using integration by parts:
Let \( u = \log_{10} x \) and \( dv = \frac{2}{x} dx \).
Then \( du = \frac{1}{x \ln 10} dx \) and \( v = 2 \log(x) \).
\(\int u \, dv = uv - \int v \, du = 2 \log_{10} x \log(x) - \int 2 \log(x) \cdot \frac{1}{x \ln 10} dx \).

The integral simplifies to:
\( 2 \log_{10} x \log(x) - \frac{2}{\ln 10} \int \frac{\log(x)}{x} dx = 2 \log_{10} x \log(x) - \frac{2}{\ln 10} \cdot \frac{\log^2(x)}{2} \),
where \(\int \frac{\log x}{x} dx = \frac{1}{2} \log^2 x\).
\( y \log_{10} x = 2 \log_{10} x \log(x) - \frac{1}{\ln 10} \log^2(x) + C \).

Substituting the initial condition \( y(e) = 0 \):
\( 0 \cdot 1 = 2 \cdot 1 \times 1 - \frac{1}{\ln 10} \cdot 1 + C \).
\( C = \frac{1}{\ln 10} - 2 \log e \), where \( \log e = 1 \).

Calculating for \( y(e^2) \):
\( y(e^2) \log_{10}(e^2) = 2 \cdot 2 \log e - \frac{\log^2(e^2)}{\ln 10} + \frac{1}{\ln 10} \).
\( y(e^2) \cdot 2 = 4 - \frac{4}{\ln 10} + \frac{1}{\ln 10} \).
\( y(e^2) = \frac{4 - 3/\ln 10}{2} \).
\( y(e^2) = \frac{3}{2} \).

Thus, \( y(e^2) = \frac{3}{2} \).