Question

If $y = y(x)$ satisfies
$(1+x^2)\frac{dy}{dx} + (2 - \tan^{-1}x) = 0$
and $y(0) = 0$, then the value of $y(1)$ is:

• A. $\frac{\pi^2}{32}$
• B. $\frac{\pi^2}{32} - \frac{\pi}{4}$
• C. $\frac{\pi}{4} - \frac{\pi^2}{32}$
• D. $\frac{\pi^2}{16}$

Hint:

When faced with an integration involving $\tan^{-1}x$ and $\frac{1}{1+x^2}$, immediately think of substitution. Since the derivative of $\tan^{-1}x$ is $\frac{1}{1+x^2}$, setting $u=\tan^{-1}x$ is almost always the correct approach.

Answer 1

The Correct Option is B

Note: There appears to be a typo in the question. The differential equation as written leads to an answer of $\frac{\pi^2}{32} - \frac{\pi}{2}$, which is not among the options. Assuming the term was $(1 - \tan^{-1}x)$ instead of $(2 - \tan^{-1}x)$ leads exactly to option (B). We will solve using this assumed correction.
Step 1: Understanding the Question:
We are given a first-order ordinary differential equation and an initial condition. We need to solve this initial value problem to find the specific solution $y(x)$ and then evaluate it at $x=1$.
Step 2: Key Formula or Approach:
The given differential equation is of the variable separable type. We will rearrange the equation to separate the variables (all $y$ terms with $dy$ and all $x$ terms with $dx$) and then integrate both sides.
Step 3: Detailed Explanation:
Let's assume the corrected differential equation is:
\[ (1+x^2)\frac{dy}{dx} + (1 - \tan^{-1}x) = 0 \] Rearrange the equation to isolate $\frac{dy}{dx}$:
\[ (1+x^2)\frac{dy}{dx} = -(1 - \tan^{-1}x) = \tan^{-1}x - 1 \] \[ \frac{dy}{dx} = \frac{\tan^{-1}x - 1}{1+x^2} \] Now, separate the variables:
\[ dy = \left( \frac{\tan^{-1}x - 1}{1+x^2} \right) dx \] Integrate both sides:
\[ \int dy = \int \left( \frac{\tan^{-1}x}{1+x^2} - \frac{1}{1+x^2} \right) dx \] \[ y(x) = \int \frac{\tan^{-1}x}{1+x^2} dx - \int \frac{1}{1+x^2} dx \] For the first integral, use the substitution $u = \tan^{-1}x$, so $du = \frac{1}{1+x^2}dx$. The integral becomes $\int u du = \frac{u^2}{2} = \frac{(\tan^{-1}x)^2}{2}$.
The second integral is a standard form: $\int \frac{1}{1+x^2}dx = \tan^{-1}x$.
So, the general solution is:
\[ y(x) = \frac{(\tan^{-1}x)^2}{2} - \tan^{-1}x + C \] Now, apply the initial condition $y(0)=0$:
\[ 0 = \frac{(\tan^{-1}0)^2}{2} - \tan^{-1}0 + C \implies 0 = 0 - 0 + C \implies C = 0 \] The particular solution is:
\[ y(x) = \frac{(\tan^{-1}x)^2}{2} - \tan^{-1}x \] Finally, evaluate the solution at $x=1$:
\[ y(1) = \frac{(\tan^{-1}1)^2}{2} - \tan^{-1}1 \] We know that $\tan^{-1}1 = \frac{\pi}{4}$.
\[ y(1) = \frac{(\pi/4)^2}{2} - \frac{\pi}{4} = \frac{\pi^2/16}{2} - \frac{\pi}{4} = \frac{\pi^2}{32} - \frac{\pi}{4} \] Step 4: Final Answer:
Assuming the typo correction, the value of $y(1)$ is $\frac{\pi^2}{32} - \frac{\pi}{4}$.