Let $𝑦=𝑦(𝑥)$ be the solution curve of the differential equation

$\left(1+𝑥^2 ight)d𝑦+\left(𝑦- an^{-1}𝑥 ight)d𝑥=0$, $𝑦(0)=1$. Then the value of $𝑦(1)$ is :

Question

Let $𝑦=𝑦(𝑥)$ be the solution curve of the differential equation

$\left(1+𝑥^2ight)d𝑦+\left(𝑦-	an^{-1}𝑥ight)d𝑥=0$, $𝑦(0)=1$. Then the value of $𝑦(1)$ is :

Accepted Answer

D. 1. **Convert to standard linear form** by dividing throughout by $(1+𝑥^2)$:
$$
\frac{d𝑦}{d𝑥}+\frac{1}{1+𝑥^2}𝑦=\frac{	an^{-1}𝑥}{1+𝑥^2}
$$

2. **Find the integrating factor**, since $𝑃(𝑥)=\frac{1}{1+𝑥^2}$ integrates to $	an^{-1}𝑥$:
$$
\mu(𝑥)=𝑒^{\int \frac{1}{1+𝑥^2}d𝑥}=𝑒^{	an^{-1}𝑥}
$$

3. **Multiply by the integrating factor**, converting the left side into a perfect derivative:
$$
\frac{d}{d𝑥}\left(𝑦\,𝑒^{	an^{-1}𝑥}ight)=𝑒^{	an^{-1}𝑥}\cdot\frac{	an^{-1}𝑥}{1+𝑥^2}
$$

4. **Substitute $𝑡=	an^{-1}𝑥$**, so $d𝑡=\frac{d𝑥}{1+𝑥^2}$, turning the right side into $\int 𝑡𝑒^{𝑡}\,d𝑡$, evaluated using integration by parts:
$$
𝑦\,𝑒^{𝑡}=𝑒^{𝑡}(𝑡-1)+𝐶
$$

5. **Rewrite in terms of $𝑥$** and isolate $𝑦$:
$$
𝑦=𝑡-1+𝐶𝑒^{-𝑡}=	an^{-1}𝑥-1+𝐶𝑒^{-	an^{-1}𝑥}
$$

6. **Apply $𝑦(0)=1$**, using $	an^{-1}0=0$:
$$
1=0-1+𝐶 \implies 𝐶=2
$$

7. **Substitute $𝑥=1$**, using $	an^{-1}1=\frac{\pi}{4}$:
$$
𝑦(1)=\frac{\pi}{4}-1+2𝑒^{-\pi/4}=\frac{2}{𝑒^{\pi/4}}+\frac{\pi}{4}-1
$$

Hence, the answer is **Option D: $\frac{2}{𝑒^{\pi/4}}+\frac{\pi}{4}-1$**.

Let 𝑦=𝑦(𝑥)𝑦=𝑦(𝑥)y=y(x) be the solution curve of the differential equation (1+𝑥2)d𝑦+(𝑦−tan⁡−1𝑥)d𝑥=0\left(1+𝑥^2\right)d𝑦+\left(𝑦-\tan^{-1}𝑥\right)d𝑥=0(1+x2)dy+(y−tan−1x)dx=0, 𝑦(0)=1𝑦(0)=1y(0)=1. Then the value of 𝑦(1)𝑦(1)y(1) is :

Let $𝑦=𝑦(𝑥)$ be the solution curve of the differential equation

$\left(1+𝑥^2\right)d𝑦+\left(𝑦-\tan^{-1}𝑥\right)d𝑥=0$ , $𝑦(0)=1$ . Then the value of $𝑦(1)$ is :