If y = y(x) and (1 + x²)dy + (1 − tan⁻¹x)dx = 0 and y(0) = 1, then y(1) is

The correct answer is:
π²/32 - π/4 + 1

Step-by-Step Explanation:

We are given the first-order differential equation:
$(1\; +\; x2)\; dy\; +\; (1\; -\; tan-1x)\; dx\; =\; 0$
with the initial condition $y(0)\; =\; 1$. We need to find the value of $y(1)$.

Step 1: Separate the variables
We can rewrite the differential equation to separate the variables $y$ and $x$:
$(1\; +\; x2)\; dy\; =\; -\; (1\; -\; tan-1x)\; dx$
Divide both sides by $1\; +\; x2$ to isolate $dy$:
$dy\; =\; -\frac{1-{\tan }^{-1}x}{1+x^2}dx$

Step 2: Integrate both sides
Integrating both sides of the equation gives:
$\int \mathrm{dy}=-\int \frac{1-{\tan }^{-1}x}{1+x^2}\mathrm{dx}$
This simplifies to:
$y=-\int \frac{1}{1+x^2}\mathrm{dx}+\int \frac{{\tan }^{-1}x}{1+x^2}\mathrm{dx}$

Step 3: Evaluate the integrals
The first integral is standard:
$\int \frac{1}{1+x^2}\mathrm{dx}={\tan }^{-1}x$
For the second integral, let's use substitution. Let:
$t\; =\; tan-1x$
Then, differentiating both sides gives:
$dt\; =\frac{1}{1+x^2}dx$
Substituting these into the second integral:
$\int \frac{{\tan }^{-1}x}{1+x^2}\mathrm{dx}=\int t\mathrm{dt}=\frac{t^2}{2}+C=\frac{{\left({\tan }^{-1}x\right)}^2}{2}+C$
where $C$ is the constant of integration.

Combining these results, the general solution is:
$y(x)\; =\; -\; tan-1x\; +\frac{{\left({\tan }^{-1}x\right)}^2}{2}+\; C$

Step 4: Apply the initial condition to find C
We are given $y(0)\; =\; 1$. Substituting $x\; =\; 0$ and $y\; =\; 1$ into the general solution:
$1\; =\; -\; tan-1(0)\; +\frac{{\left({\tan }^{-1}\right(0\left)\right)}^2}{2}+\; C$
Since $tan-1(0)\; =\; 0$, we get:
$1\; =\; 0\; +\; 0\; +\; C\; \Rightarrow \; C\; =\; 1$
Therefore, the particular solution is:
$y(x)\; =\; -\; tan-1x\; +\frac{{\left({\tan }^{-1}x\right)}^2}{2}+\; 1$

Step 5: Calculate y(1)
Substitute $x\; =\; 1$ into the particular solution:
$y(1)\; =\; -\; tan-1(1)\; +\frac{{\left({\tan }^{-1}\right(1\left)\right)}^2}{2}+\; 1$
We know that $tan-1(1)\; =\frac{\pi }{4}$. Substituting this value:
$y(1)\; =\; -\frac{\pi }{4}+\frac{1}{2}{\left(\frac{\pi }{4}\right)}^2+\; 1$
$y(1) = - π4 + 12 (π216) y(1) = - π4 + π232 + 1 Rearranging the terms: y(1) = π232 - π4 + 1 Unlock Our Free Library Access expert-curated educational resources and study materials—completely free. Explore Get 3 Months FREE trial Create, conduct, and manage professional online assessments with Mindyard. Perfect for teachers and institutes. Claim Discover more resources You may also like Study Material View All 150 Questions CTET CTET Paper 2 Set E 20th Aug 2023 Question Paper with Solutions By Gaurav Kumar View Paper 150 Questions CTET CTET Paper 2 Set M 31st Jan 2021 Question Paper with Solutions By Gaurav Kumar View Paper Mock Tests View All CTET intermediate No time limit child development and pedagogy, mathematics, social science CTET Paper 2 Set E 20th Aug 2023 By Gaurav Kumar Start Test SSC intermediate 2 hours and 30 mins child development and pedagogy, mathematics, social science CTET Paper 2 Set M 31st Jan 2021 By Gaurav Kumar Start Test Confirm Logout Are you sure you want to log out of your account? Cancel Logout Report Issue What's wrong with this content? Select an issue type Sexual content Copyright Violation Violent or repulsive content Hateful or abusive content Other If you report a Copyright Violation, you will be contacted at your registered email for further details and verification. Question Number (e.g., 2, 5, 12-18) Comment / Explanation Your Contact (Email / Mobile) Submit Report Save to revision list Create New Revision List Cancel Create Login Required Log In Sign up Login Required You need to log in first to report this item. Log In Sign up Copyright © 2026 Mindyard. All Rights Reserved. Terms of Service Privacy Policy User Agreement Refund Policy Shipping & Delivery Login Ask AI Tutor 5 left Q1 View Question & Options AI Tutor is solving this question... Reading question context & options... Simpler terms Why correct? Shortcut Common traps Maximum follow-ups reached for this question. $