Q.(xi) The integrating factor of $\frac{dy}{dx}+y=\frac{1+y}{x}$ is ______.

The problem asks us to identify the function $f(y)$ by comparing the general solution of a given differential equation with a specified form. This requires us to solve the differential equation first. The key concept here is recognizing and solving a first-order linear differential equation.

A first-order linear differential equation has the general form $\frac{dx}{dy} + P(y)x = Q(y)$ (when $x$ is the dependent variable and $y$ is the independent variable). The "why" behind this method is that multiplying the entire equation by a special function, called the integrating factor, transforms the left-hand side into the exact derivative of a product, making the equation directly integrable.

Here's how we solve it:

1. Rearrange the differential equation into standard linear form:

   The given differential equation is $(1 + y^2) dx = (\tan^{-1} y - x) dy$.

   To get it into the form $\frac{dx}{dy} + P(y)x = Q(y)$, we first divide by $dy$:

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

   Now, move the term involving $x$ to the left side:

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

   Finally, divide by $(1 + y^2)$ to isolate $\frac{dx}{dy}$:

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

2. Identify $P(y)$ and $Q(y)$:

   Comparing our rearranged equation with the standard form $\frac{dx}{dy} + P(y)x = Q(y)$, we have:

   $P(y) = \frac{1}{1 + y^2}$

   $Q(y) = \frac{\tan^{-1} y}{1 + y^2}$

3. Calculate the Integrating Factor (IF):

   The integrating factor for $\frac{dx}{dy} + P(y)x = Q(y)$ is $IF = e^{\int P(y) dy}$.

   Substitute $P(y)$:

   $IF = e^{\int \frac{1}{1 + y^2} dy}$

   We know that $\int \frac{1}{1 + y^2} dy = \tan^{-1} y$.

   So, the integrating factor is:

   $IF = e^{\tan^{-1} y}$

4. Apply the general solution formula:

   The general solution of a first-order linear differential equation is $x \cdot IF = \int (Q(y) \cdot IF) dy + c$.

   Substitute $IF$ and $Q(y)$ into this formula:

   $x \cdot e^{\tan^{-1} y} = \int \left( \frac{\tan^{-1} y}{1 + y^2} \cdot e^{\tan^{-1} y} \right) dy + c$

5. Evaluate the integral:

   The integral on the right-hand side is $\int \frac{\tan^{-1} y}{1 + y^2} e^{\tan^{-1} y} dy$.

   This integral can be solved using substitution and integration by parts.

   Let $t = \tan^{-1} y$.

   Then, differentiating both sides with respect to $y$, we get $dt = \frac{1}{1 + y^2} dy$. …

The problem asks for the general solution of a given differential equation. This equation is of the form $M(x,y)\,dx + N(x,y)\,dy = 0$. Our strategy will be to first check if it's an exact differential equation. If not, we look for an integrating factor that can transform it into an exact (or even separable) equation, which can then be solved by integration.

Here's a step-by-step approach:

1. Identify $M$ and $N$:

   The given differential equation is:

   $((1+x^2)y\sin x - 2xy)\,dx - \log y(1+x^2)\,dy = 0$

   Comparing this with the standard form $M\,dx + N\,dy = 0$, we identify:

   $M = (1+x^2)y\sin x - 2xy$

   $N = -\log y(1+x^2)$

2. Check for exactness:

   An equation is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Let's compute these partial derivatives:

   $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}((1+x^2)y\sin x - 2xy) = (1+x^2)\sin x - 2x$

   $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(-\log y(1+x^2)) = -\log y(2x)$

   Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the given differential equation is not exact.

3. Find an integrating factor by observation:

   Since the equation is not exact, we look for an integrating factor. Sometimes, a simple division by a common factor can simplify the equation significantly. Let's examine the terms in $M$ and $N$:

   $M = y((1+x^2)\sin x - 2x)$

   $N = -(1+x^2)\log y$

   Notice that $y$ is a common factor in $M$, and $(1+x^2)$ is a common factor in $N$. This suggests that dividing the entire equation by $y(1+x^2)$ might simplify it. Let's try using $\mu = \frac{1}{y(1+x^2)}$ as an integrating factor.

   Multiplying the original equation by $\mu$:

   $\frac{(1+x^2)y\sin x - 2xy}{y(1+x^2)}\,dx - \frac{\log y(1+x^2)}{y(1+x^2)}\,dy = 0$

   Simplify each term:

   $\left(\frac{(1+x^2)y\sin x}{y(1+x^2)} - \frac{2xy}{y(1+x^2)}\right)\,dx - \frac{\log y}{y}\,dy = 0$

   This simplifies to:

   $\left(\sin x - \frac{2x}{1+x^2}\right)\,dx - \frac{\log y}{y}\,dy = 0$

4. Verify the new equation is separable:

   Let the new coefficients be $M'$ and $N'$:

   $M' = \sin x - \frac{2x}{1+x^2}$

   $N' = -\frac{\log y}{y}$

   Notice that $M'$ is a function of $x$ alone, and $N'$ is a function of $y$ alone. This means the equation is now separable, which is a special case of an exact equation. We can integrate each term independently.

5. Integrate the separable equation:

   Integrate both sides of the transformed equation:

   $\int \left(\sin x - \frac{2x}{1+x^2}\right)\,dx - \int \frac{\log y}{y}\,dy = C$

   Let's evaluate each integral: …

Nature of the family. The given equation is separable, and its integral has the form $\;\text{(trig }\sqrt{x})+e^{y}=C$. The required curve must pass through $(\pi^2,1)$, where $\sqrt{x}=\pi$, so $\sin\sqrt{x}=\sin\pi=0$, $\cos\sqrt{x}=\cos\pi=-1$, and $e^{y}=e^{1}=e$.

Test each candidate at $(\pi^2,1)$:

• (A) $\sin\sqrt{x}+e^{y}=1+e:\ 0+e=e\neq 1+e.$ ✗ …

Setup. The equation is

Let $v=2x+y$. Then $\dfrac{dv}{dx}=2+\dfrac{dy}{dx}$, i.e. $\dfrac{dy}{dx}=\dfrac{dv}{dx}-2$. Substituting:

Separate the variables.

$\frac{dv}{\sin v}=-\frac{dx}{\cos x}\;\Longrightarrow\;\int\csc v\,dv=-\int\sec x\,dx.$ …

The key here is recognising the structure of the equation. Every term has a total degree of 3 — $x^3 dx$, $x y^2 dy$, $y^3 dx$ — so the equation is homogeneous in $x$ and $y$. For homogeneous differential equations, the standard trick is to set $y = vx$, which turns the equation into one where variables separate cleanly.

Let’s work through it.

1. Rewrite the equation in standard form. The given equation is

Group the $dx$ terms together:

So

2. Check homogeneity. Replace $x$ by $tx$ and $y$ by $ty$ in the right-hand side:

The function is unchanged — it is homogeneous of degree 0. So the substitution $y = vx$ is valid.

3. Substitute $y = vx$. Then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. The equation becomes

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

So

5. Integrate both sides.

gives

where $C$ is the constant of integration.

6. Back-substitute $v = y/x$.

Multiply through by 3:

Let $c = 3C$ (still an arbitrary constant), so

7. Solve for $y^3$. Multiply by $x^3$:

But note: the constant $c$ here is arbitrary, so $c x^3$ can be written simply as $c x^3$ — it’s already a general constant times $x^3$. However, the given options have $y^3 = x^3 \log|x| + c$ (option B). That means the constant 3 has been absorbed into the log term? Let’s check carefully.

Actually, $3\log|x| = \log|x|^3$, so

But $c x^3$ is not the same as a plain $+c$ unless we redefine constants differently. Wait — the standard form in option (B) is $y^3 = x^3 \log|x| + c$. That suggests the constant 3 has been absorbed into the arbitrary constant. Let’s see: if we write

then $y^3 = x^3 \log|x|^3 + C x^3$. But $C x^3$ is not a constant; it’s a function of $x$. So that can’t be right.

The resolution: the constant $c$ in option (B) is actually an arbitrary constant, and the expression $y^3 = x^3 \log|x| + c$ is obtained by absorbing the factor of 3 into the constant differently. Let’s re-derive cleanly:

From $\frac{v^3}{3} = \log|x| + C$, multiply by 3:

Let $k = 3C$, so $v^3 = 3\log|x| + k$.

Then $y^3 = x^3 v^3 = x^3 (3\log|x| + k) = 3x^3 \log|x| + k x^3$.

This is not of the form $x^3 \log|x| + c$ unless we redefine the log term. But note: $3x^3 \log|x| = x^3 \log|x|^3$, and if we let $c = k$, then $y^3 = x^3 \log|x|^3 + c x^3$. That still has $c x^3$, not $+c$. …

We start with the differential equation

It is written in the form $M\,dx + N\,dy = 0$ with

Check for exactness:

Since $1 \neq 2$, the equation is not exact. However, the variables suggest we might find an integrating factor that depends only on $y$ (because the $x$-dependence appears only linearly in $N$).

1. Find an integrating factor $\mu(y)$. For an integrating factor depending only on $y$, the condition is

This is a function of $y$ alone, so

(We take the simplest positive factor; the constant of integration is absorbed into the constant later.)

2. Multiply the original equation by $\mu = y$:

Now we have new coefficients:

Check exactness again:

They match, so the equation is now exact.

3. Solve the exact equation. We need a function $F(x,y)$ such that

Integrate the first with respect to $x$:

where $g(y)$ is an unknown function of $y$.

4. Determine $g(y)$. Differentiate $F$ with respect to $y$: $\frac{\partial F}{\partial y} = 2xy + g'(y).$ …

Rewrite the equation:

Since $9x - 3y = 3(3x - y)$, put $v = 3x - y$, so $\dfrac{dy}{dx} = 3 - \dfrac{dv}{dx}$ and

Separate variables:

Split the left side: with $8v + 14 = 2(4v + 7)$,

Integrating directly,

Multiply by $32$ and use $8v + 14 = 2(12x - 4y + 7)$, $\,12v = 36x - 12y$:

$36x - 12y - \log|12x - 4y + 7| = 32x + C',$ …

Start from $2\,dx+dy=(6xy+4x-3y)\,dx$. Move $2\,dx$ to the right:

Factor the right side.

So

Separate and integrate. …

The left side is an exact derivative, since with $\sec^2 x = 1+\tan^2 x$:

Integrate the right side with $u=\tan x$, $du=\sec^2 x\,dx$, $\sec^2 x = 1+u^2$:

So the family of curves is

Applying the intended initial condition ($C=0$) and evaluating at $x=\tfrac{\pi}{3}$ where $\tan x = \sqrt{3}$: …