Q.Find the particular solution of the differential equation $\frac{dy}{dx} + y\cot x = 2x + x^2 \cot x$ $(x \neq 0)$ given that $y = 0$ when $x = \frac{\pi}{2}$.

We have the differential equation

This is a first-order linear ordinary differential equation of the form

The key idea: we multiply both sides by an integrating factor that makes the left side a perfect derivative of a product. That lets us integrate directly.

1. Identify $P(x)$ and find the integrating factor Here $P(x) = \frac{1}{x}$. The integrating factor is

(We take the simplest positive version; the constant of integration is irrelevant here.)

2. Multiply the whole equation by $\mu(x) = x$

simplifies to

3. Recognize the left side as a derivative Notice that

So the equation becomes

4. Integrate both sides

gives

$x y = \frac{x^4}{4} + c,$ …

We are given:

and told that the solution can be written as $y = x \tan(f(x)) + c$. We need to identify $f(x)$ from the options.

Concept & Intuition:

The right-hand side is a ratio of linear expressions in $x$ and $y$ where the numerator and denominator are both homogeneous of degree 1. This is a classic homogeneous differential equation. The standard trick: set $y = vx$ (so $v = y/x$), which turns the equation into one where variables separate. After integration, we expect an expression involving $\tan^{-1}(v)$ and a logarithm of $x$. The given form $y = x \tan(f(x))$ suggests that $f(x)$ is actually the angle whose tangent is $y/x$, i.e., $f(x) = \tan^{-1}(y/x)$ plus possibly a constant or a log term. Let’s work it out.

1. Rewrite using substitution $y = vx$ Let $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substitute into the DE:

2. Separate variables Subtract $v$ from both sides:

So:

Separate:

3. Integrate both sides Left side: split the fraction:

Right side:

So we have:

4. Rewrite in terms of $x$ and $y$ Recall $v = y/x$, so $1+v^2 = 1 + \frac{y^2}{x^2} = \frac{x^2 + y^2}{x^2}$. Then:

Substitute into the integrated equation:

Simplify:

$3\tan^{-1}\left(\frac{y}{x}\right) - \log(x^2 + y^2) + 2\log|x| = 2\log|x| + C.$ …

We start with

It is natural to treat $x$ as a function of $y$ because the right‑hand side mixes $x$ and $y$ in a way that suggests a linear equation in $x$. Dividing through by $dy$ and by $1+y^2$ puts it into a familiar form.

1. Rewrite as a linear ODE in $x(y)$. Divide both sides by $dy$ and by $1+y^2$:

Rearranging terms to isolate the derivative:

This is a first‑order linear ODE:

2. Find the integrating factor. The standard integrating factor is

(Recall $\int \frac{dy}{1+y^2} = \tan^{-1}y + C$; we take the simplest antiderivative.)

3. Multiply through and integrate. Multiply the ODE by $\mu(y)$:

The left‑hand side is the derivative of $x\,e^{\tan^{-1}y}$ because

So we have

4. Integrate both sides.

Let $u = \tan^{-1}y$. Then $du = \frac{dy}{1+y^2}$. The integral becomes

Integration by parts: let $w = u$, $dv = e^{u}du$ → $dw = du$, $v = e^{u}$. Then

$\int u e^{u}\,du = u e^{u} - \int e^{u}\,du = u e^{u} - e^{u} + C.$ …

Concept and intuition:

The problem asks for the angle between two curves at their intersection. That angle is defined as the acute angle between their tangent lines. So we need:

• Solve the differential equation to get the family of curves.
• Pick the specific curve passing through $(1,0)$.
• Find the intersection point(s) of that curve with the ellipse $x^2+3y^2=3$.
• Compute the slopes of both curves at that point.
• Use the formula $\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ to get $\theta$.

Step-by-step solution:

1. Solve the differential equation Given: $x y \, dy - (1 + y^2) dx = 0$. Rearranging: $x y \, dy = (1 + y^2) dx$. Separate variables:

Integrate both sides:

The left integral: let $u = 1+y^2$, $du = 2y\,dy$, so $\frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\log|1+y^2|$.

The right: $\log|x| + C$.

So:

Multiply by 2: $\log(1+y^2) = 2\log|x| + 2C = \log(x^2) + \log C_1$ (let $C_1 = e^{2C}$).

Hence:

This is the family of curves.

2. Apply the initial condition $(1,0)$ Plug $x=1$, $y=0$: $1 + 0 = C_1 \cdot 1 \implies C_1 = 1$. So the specific curve is:

3. Find intersection with the ellipse $x^2 + 3y^2 = 3$ Substitute $y^2 = x^2 - 1$ into the ellipse:

Then $y^2 = x^2 - 1 = \frac{3}{2} - 1 = \frac{1}{2}$.

So intersection points: $\left(\pm\sqrt{\frac{3}{2}}, \pm\sqrt{\frac{1}{2}}\right)$.

By symmetry, the angle will be the same at all intersections; pick the first quadrant:

4. Find slopes at the intersection
   • For curve $y^2 = x^2 - 1$: differentiate implicitly: $2y y' = 2x \implies y' = \frac{x}{y}$. At $\left(\sqrt{\frac{3}{2}},\ \sqrt{\frac{1}{2}}\right)$: …

The core idea is that when $x$ and $y$ are both given as functions of a parameter $\theta$, the derivative $\frac{dy}{dx}$ can be written as $\frac{dy/d\theta}{dx/d\theta}$. Here we are told $\frac{dy}{dx}$ directly in terms of $\theta$, so we can find $dx/d\theta$ from the given $x(\theta)$, then solve for $dy/d\theta$ and integrate to get $y = f(\theta)$. The condition $f(2\pi)=0$ determines the integration constant.

1. Find $dx/d\theta$. Given $x = a(\cos\theta + \theta\sin\theta)$, differentiate with respect to $\theta$:

The $\sin\theta$ terms cancel neatly — that’s the design of the expression.

2. Use the chain rule to get $dy/d\theta$. We know $\frac{dy}{dx} = \frac{\tan\theta}{\theta}$. Since $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$, we have

Multiply through:

So $dy/d\theta = a\sin\theta$.

3. Integrate to find $f(\theta)$. $y = \int a\sin\theta\,d\theta = -a\cos\theta + C,$ …

The key insight is that the right-hand side $(3x + y + 4)^2$ depends on $x$ and $y$ only through the combination $3x + y + 4$. This suggests a substitution that collapses the two variables into one, turning the equation into a separable first-order ODE.

1. Set up the substitution. Let $u = 3x + y + 4$. Then differentiate with respect to $x$:

The given equation is $\frac{dy}{dx} = u^2$, so

2. Separate variables. We now have

Integrate both sides:

3. Evaluate the integrals. Recall $\int \frac{du}{a^2 + u^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$. Here $a^2 = 3$, so $a = \sqrt{3}$. Thus

4. Back-substitute $u$. Since $u = 3x + y + 4$, we get $\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{3x + y + 4}{\sqrt{3}}\right) - x = C.$ …

The problem asks us to evaluate a definite integral and then use its structure to identify an unknown function $f(x)$. The key to solving this type of problem is to choose an effective substitution that simplifies the integrand. The presence of $\sqrt{e^x + 4}$ in the denominator and $e^x$ in the numerator strongly suggests a substitution involving $e^x$ or the entire square root term. Substituting $t = \sqrt{e^x + 4}$ is a powerful technique here, as it helps to rationalize the denominator and transform the exponential terms into simpler polynomial forms, making the integration straightforward. Once the integral is evaluated, we compare it with the given form to deduce $f(x)$ and then use the provided condition to find the required value.

1. Choose an appropriate substitution:

   The integral is $\int \frac{e^x + 3}{\sqrt{e^x + 4}} dx$.

   Let's simplify the expression by substituting the square root term.

   Let $t = \sqrt{e^x + 4}$.

2. Express all parts of the integral in terms of $t$ and $dt$:

   From $t = \sqrt{e^x + 4}$, we square both sides:

   $t^2 = e^x + 4$.

   This allows us to express $e^x$ in terms of $t$:

   $e^x = t^2 - 4$.

   Now, differentiate $t^2 = e^x + 4$ with respect to $x$:

   $2t \frac{dt}{dx} = e^x$.

   Rearranging to find $dx$:

   $dx = \frac{2t}{e^x} dt$.

   Substitute $e^x = t^2 - 4$ into the expression for $dx$:

   $dx = \frac{2t}{t^2 - 4} dt$.

   Next, express the numerator $e^x + 3$ in terms of $t$:

   $e^x + 3 = (t^2 - 4) + 3 = t^2 - 1$.

   The denominator $\sqrt{e^x + 4}$ is simply $t$.

3. Rewrite and simplify the integral using the substitution:

   Substitute these expressions back into the original integral:

The $t$ in the denominator and numerator cancels out:

To integrate this rational function, we can perform polynomial long division or manipulate the numerator to match the denominator:

4. Evaluate the simplified integral: The first part is straightforward: $\int 2 \, dt = 2t$. For the second part, we use the standard integral formula for $\frac{1}{x^2 - a^2}$:

   $\int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C$

   In our case, $x=t$ and $a=2$ (since $4 = 2^2$).

Combining both parts, the integral evaluates to: …