Question

Let \(y=y(x)\) be the solution curve of the differential equation \((1+x^2)dy+(y-\tan^{-1}x) dx=0\), \(y(0) = 1\). Then the value of \(y(1)\) is:

• A. \(\frac{4}{e^{\pi/4}} - \frac{\pi}{2} - 1\)
• B. \(\frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1\)
• C. \(\frac{2}{e^{\pi/4}} - \frac{\pi}{4} - 1\)
• D. \(\frac{4}{e^{\pi/4}} + \frac{\pi}{2} - 1\)

Hint:

Recognizing the standard form of a differential equation is the most crucial step. For first-order equations, check if they are separable, homogeneous, linear, or exact. Once identified as linear (\(\frac{dy}{dx} + P(x)y = Q(x)\)), the integrating factor method is a straightforward algorithm to follow.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are asked to solve a first-order differential equation with an initial condition, which is an initial value problem. After finding the particular solution, we need to evaluate it at \(x=1\).
Step 2: Key Formula or Approach:
The given differential equation can be arranged into the standard linear form: \(\frac{dy}{dx} + P(x)y = Q(x)\). The solution is given by \(y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C\), where the integrating factor (I.F.) is \(e^{\int P(x) dx}\).
Step 3: Detailed Explanation:
The given equation is \((1+x^2)dy+(y-\tan^{-1}x) dx=0\). Rearrange it to find \(\frac{dy}{dx}\): \[ (1+x^2)\frac{dy}{dx} + y - \tan^{-1}x = 0 \] \[ (1+x^2)\frac{dy}{dx} + y = \tan^{-1}x \] Divide by \((1+x^2)\) to get the standard linear form: \[ \frac{dy}{dx} + \frac{1}{1+x^2}y = \frac{\tan^{-1}x}{1+x^2} \] This is in the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = \frac{1}{1+x^2}\) and \(Q(x) = \frac{\tan^{-1}x}{1+x^2}\). First, calculate the integrating factor (I.F.): \[ \text{I.F.} = e^{\int P(x) dx} = e^{\int \frac{1}{1+x^2} dx} = e^{\tan^{-1}x} \] The general solution is: \[ y \cdot e^{\tan^{-1}x} = \int \frac{\tan^{-1}x}{1+x^2} \cdot e^{\tan^{-1}x} dx + C \] To solve the integral on the right, let \(t = \tan^{-1}x\). Then \(dt = \frac{1}{1+x^2} dx\). The integral becomes \(\int t e^t dt\). We solve this using integration by parts (\(\int u dv = uv - \int v du\)). Let \(u=t\) and \(dv = e^t dt\). Then \(du=dt\) and \(v=e^t\). \[ \int t e^t dt = t e^t - \int e^t dt = t e^t - e^t = (t-1)e^t \] Substitute back \(t = \tan^{-1}x\): \[ y \cdot e^{\tan^{-1}x} = (\tan^{-1}x - 1)e^{\tan^{-1}x} + C \] Divide by \(e^{\tan^{-1}x}\): \[ y(x) = (\tan^{-1}x - 1) + C e^{-\tan^{-1}x} \] Now use the initial condition \(y(0)=1\) to find C. \[ 1 = (\tan^{-1}(0) - 1) + C e^{-\tan^{-1}(0)} \] \[ 1 = (0 - 1) + C e^0 \implies 1 = -1 + C \implies C = 2 \] The particular solution is: \[ y(x) = \tan^{-1}x - 1 + 2e^{-\tan^{-1}x} \] Finally, we need to find \(y(1)\): \[ y(1) = \tan^{-1}(1) - 1 + 2e^{-\tan^{-1}(1)} \] \[ y(1) = \frac{\pi}{4} - 1 + 2e^{-\pi/4} \] \[ y(1) = \frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1 \] Step 4: Final Answer:
The value of \(y(1)\) is \(\frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1\).