Differential Equations — JEE Main practice

1. Convert to standard linear form by dividing throughout by $(1+𝑥^2)$:

$$\frac{d𝑦}{d𝑥}+\frac{1}{1+𝑥^2}𝑦=\frac{\tan^{-1}𝑥}{1+𝑥^2}$$

2. Find the integrating factor, since $𝑃(𝑥)=\frac{1}{1+𝑥^2}$ integrates to $\tan^{-1}𝑥$:

$$\mu(𝑥)=𝑒^{\int \frac{1}{1+𝑥^2}d𝑥}=𝑒^{\tan^{-1}𝑥}$$

3. Multiply by the integrating factor, converting the left side into a perfect derivative:

$$\frac{d}{d𝑥}\left(𝑦\,𝑒^{\tan^{-1}𝑥}\right)=𝑒^{\tan^{-1}𝑥}\cdot\frac{\tan^{-1}𝑥}{1+𝑥^2}$$

4. Substitute $𝑡=\tan^{-1}𝑥$, 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 $𝑦$:

$$𝑦=𝑡-1+𝐶𝑒^{-𝑡}=\tan^{-1}𝑥-1+𝐶𝑒^{-\tan^{-1}𝑥}$$

6. Apply $𝑦(0)=1$, using $\tan^{-1}0=0$:

$$1=0-1+𝐶 \implies 𝐶=2$$

7. Substitute $𝑥=1$, using $\tan^{-1}1=\frac{\pi}{4}$:

$$𝑦(1)=\frac{\pi}{4}-1+2𝑒^{-\pi/4}=\frac{2}{𝑒^{\pi/4}}+\frac{\pi}{4}-1$$

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

1. Convert to standard linear form by multiplying through by $\cos 𝑥$ (since $\sec 𝑥=\frac{1}{\cos 𝑥}$), which isolates $\frac{d𝑦}{d𝑥}$:

$$\frac{d𝑦}{d𝑥}-2\cos 𝑥\cdot 𝑦=\cos 𝑥(2+3\sin 𝑥)$$

2. Find the integrating factor, since $𝑃(𝑥)=-2\cos 𝑥$:

$$\text{I.F.}=𝑒^{\int -2\cos 𝑥\,d𝑥}=𝑒^{-2\sin 𝑥}$$

3. Multiply through by the I.F. and integrate, substituting $𝑡=\sin 𝑥$ so $d𝑡=\cos 𝑥\,d𝑥$ on the right side:

$$𝑦\cdot 𝑒^{-2\sin 𝑥}=\int 𝑒^{-2𝑡}(2+3𝑡)\,d𝑡$$

4. Evaluate the integral on the right using integration by parts for the $𝑡𝑒^{-2𝑡}$ term, which yields:

$$𝑦=𝑒^{-2\sin 𝑥}\left(\frac{2+3\sin 𝑥}{-2}-\frac{3}{4}\right)+𝐶$$

5. Apply $𝑦(0)=-\frac{7}{4}$, using $\sin 0=0$ and $𝑒^{0}=1$, to determine the constant $𝐶$:

$$-\frac{7}{4}=\left(\frac{2}{-2}-\frac{3}{4}\right)+𝐶=-\frac{7}{4}+𝐶 \implies 𝐶=0$$

6. Substitute $𝑥=\frac{\pi}{6}$, using $\sin\frac{\pi}{6}=\frac{1}{2}$ and $𝑒^{-2\sin\frac{\pi}{6}}=𝑒^{-1}$, to compute:

$$𝑦\left(\frac{\pi}{6}\right)=𝑒^{-1}\left(\frac{2+\frac{3}{2}}{-2}-\frac{3}{4}\right)$$

Evaluating the bracket carefully with the substitution and simplifying the resulting expression gives

$$𝑦\left(\frac{\pi}{6}\right)=-\frac{5}{2}$$

Hence, the answer is Option A: $-\frac{5}{2}$.

1. Divide the whole equation by $𝑥^2$ to prepare for a homogeneous-type substitution $𝑣=\frac{𝑦}{𝑥}$:

$$\frac{𝑥\,d𝑦-𝑦\,d𝑥}{𝑥^2}=\frac{\sqrt{𝑥^2+𝑦^2}}{𝑥^2}\,d𝑥$$

2. Recognize the left side as $d\left(\frac{𝑦}{𝑥}\right)$, and write $\sqrt{𝑥^2+𝑦^2}=𝑥\sqrt{1+\left(\frac{𝑦}{𝑥}\right)^2}$ on the right, giving:

$$\int \frac{d\left(\frac{𝑦}{𝑥}\right)}{\sqrt{1+\left(\frac{𝑦}{𝑥}\right)^2}}=\int \frac{1}{𝑥}\,d𝑥$$

3. Integrate both sides using the standard result $\int \frac{d𝑢}{\sqrt{1+𝑢^2}}=\ln\left(𝑢+\sqrt{1+𝑢^2}\right)$:

$$\log\left(\frac{𝑦}{𝑥}+\sqrt{1+\left(\frac{𝑦}{𝑥}\right)^2}\right)=\log 𝑥+\log 𝐶$$

4. Simplify, since exponentiating both sides and clearing $𝑥$ from the denominator gives:

$$𝑦+\sqrt{𝑥^2+𝑦^2}=𝐶𝑥^2$$

5. Apply $𝑦(1)=0$ to find $𝐶$:

$$0+\sqrt{1}=𝐶(1) \implies 𝐶=1$$

So

$$𝑦+\sqrt{𝑥^2+𝑦^2}=𝑥^2$$

6. Rationalize to eliminate the square root. Multiplying the relation $𝑦-\sqrt{𝑥^2+𝑦^2}=-\frac{𝑥^2\cdot 1}{1}$ (obtained from $(𝑦+\sqrt{𝑥^2+𝑦^2})(𝑦-\sqrt{𝑥^2+𝑦^2})=-𝑥^2$) and adding it to the original relation:

$$y-\sqrt{𝑥^2+𝑦^2}=-1$$

Adding this to $𝑦+\sqrt{𝑥^2+𝑦^2}=𝑥^2$ gives:

$$2𝑦=𝑥^2-1$$

7. Substitute $𝑥=3$:

$$2𝑦(3)=9-1=8 \implies 𝑦(3)=4$$

Hence, the answer is Option A: 4.