Question

Let \(y(x)\) be the solution of the differential equation \(2x^2 dy + (e^y - 2x)dx = 0, x>0\). If \(y(e)=1\), then \(y(1)\) is equal to :

• A. 2
• B. \(\log_e(2e)\)
• C. \(\log_e(2)\)
• D. 0

Hint:

When a differential equation is not immediately recognizable as linear, separable, or homogeneous, try to rearrange it. The presence of a term like \(e^y\) often suggests a substitution to simplify the equation. Transforming a Bernoulli-type equation into a linear one is a standard and powerful technique.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a first-order differential equation with an initial condition. Our goal is to find the particular solution and then evaluate it at \(x=1\).
Step 2: Key Formula or Approach:
The given differential equation is:
\[ 2x^2 dy + (e^y - 2x)dx = 0 \] Let's rearrange it to identify its type.
\[ 2x^2 \frac{dy}{dx} + e^y - 2x = 0 \] \[ \frac{dy}{dx} + \frac{1}{2x^2}e^y = \frac{1}{x} \] This equation is a form of Bernoulli's differential equation. We can convert it to a linear differential equation by making a suitable substitution.
Multiply the equation by \(e^{-y}\):
\[ e^{-y}\frac{dy}{dx} + \frac{1}{2x^2} = \frac{1}{x}e^{-y} \] Let \(z = e^{-y}\). Differentiating with respect to x, we get \( \frac{dz}{dx} = -e^{-y}\frac{dy}{dx} \).
So, \( e^{-y}\frac{dy}{dx} = -\frac{dz}{dx} \).
Substituting this into the transformed equation:
\[ -\frac{dz}{dx} + \frac{1}{2x^2} = \frac{1}{x}z \] \[ \frac{dz}{dx} + \frac{1}{x}z = \frac{1}{2x^2} \] This is a linear differential equation of the form \( \frac{dz}{dx} + P(x)z = Q(x) \), where \( P(x) = \frac{1}{x} \) and \( Q(x) = \frac{1}{2x^2} \). The solution is found using an integrating factor (I.F.).
I.F. = \( e^{\int P(x)dx} \)
Solution: \( z \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C \)
Step 3: Detailed Explanation:
First, calculate the integrating factor:
\[ \text{I.F.} = e^{\int \frac{1}{x}dx} = e^{\ln x} = x \] Now, find the solution for z:
\[ z \cdot x = \int \frac{1}{2x^2} \cdot x \, dx + C \] \[ zx = \int \frac{1}{2x} \, dx + C \] \[ zx = \frac{1}{2}\ln|x| + C \] Since the problem states \(x>0\), we have \(zx = \frac{1}{2}\ln(x) + C\).
Substitute back \(z = e^{-y}\):
\[ x e^{-y} = \frac{1}{2}\ln(x) + C \] Now, use the given condition \(y(e) = 1\) to find the constant C. Substitute \(x=e\) and \(y=1\):
\[ e \cdot e^{-1} = \frac{1}{2}\ln(e) + C \] \[ 1 = \frac{1}{2}(1) + C \] \[ C = 1 - \frac{1}{2} = \frac{1}{2} \] The particular solution is:
\[ x e^{-y} = \frac{1}{2}\ln(x) + \frac{1}{2} \] We need to find the value of \(y(1)\), so we substitute \(x=1\):
\[ 1 \cdot e^{-y} = \frac{1}{2}\ln(1) + \frac{1}{2} \] \[ e^{-y} = \frac{1}{2}(0) + \frac{1}{2} \] \[ e^{-y} = \frac{1}{2} \] Taking the reciprocal of both sides:
\[ e^y = 2 \] Taking the natural logarithm of both sides:
\[ y = \ln(2) = \log_e(2) \] Step 4: Final Answer:
The value of \(y(1)\) is \(\log_e(2)\). This corresponds to option (C).