The function $y = f(x)$ is the solution of the differential equation $\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^4 + 2x}{\sqrt{1-x^2}}$ in $(-1, 1)$ satisfying $f(0) = 0$. Then $\int_{-\sqrt{3}/2}^{\sqrt{3}/2} f(x) dx$ is Options: (A) $\frac{\pi}{3} - \frac{\sqrt{3}}{2}$ (B) $\frac{\pi}{3} - \frac{\sqrt{3}}{4}$ (C) $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$ (D) $\frac{\pi}{6} - \frac{\sqrt{3}}{2}$

### Analyzing the Setup The given differential equation is: $$\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^4 + 2x}{\sqrt{1 - x^2}}$$ This expression follows the standard form of a **Linear Differential Equation**: $\frac{dy}{dx} + P(x)y = Q(x)$. By inspection, we identify: $$P(x) = \frac{x}{x^2 - 1}, \quad Q(x) = \frac{x^4 + 2x}{\sqrt{1 - x^2}}$$ ### Determining the Integrating Factor The Integrating Factor ($I.F.$) is defined as $e^{\int P(x) dx}$. We compute the integral: $$\int P(x) dx = \int \frac{x}{x^2 - 1} dx = \frac{1}{2} \ln|x^2 - 1|$$ Given the domain $x \in (-1, 1)$, we note that $x^2 < 1$, which implies $x^2 - 1 < 0$. Therefore, $|x^2 - 1| = 1 - x^2$. The $I.F.$ becomes: $$I.F. = e^{\ln\sqrt{1 - x^2}} = \sqrt{1 - x^2}$$ ### Solving the Differential Equation Multiplying the original differential equation by the $I.F.$ simplifies the expression significantly. The equation becomes: $$\frac{d}{dx} \left( y \cdot \sqrt{1 - x^2} \right) = x^4 + 2x$$ Integrating both sides with respect to $x$: $$y \sqrt{1 - x^2} = \int (x^4 + 2x) dx = \frac{x^5}{5} + x^2 + C$$ Using the initial condition $f(0) = 0$, we find $C = 0$. Thus, the explicit function is: $$f(x) = \frac{\frac{x^5}{5} + x^2}{\sqrt{1 - x^2}}$$ ### Final Calculation via Symmetry We are tasked with evaluating the definite integral $I = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$. We split the integral into two parts: $$I = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^5}{5\sqrt{1 - x^2}} dx + \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1 - x^2}} dx$$ The first term is an **odd function** integrated over symmetric limits, which evaluates to $0$. The second term is an **even function**, allowing us to simplify: $$I = 2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1 - x^2}} dx$$ Using the substitution $x = \sin\theta$, where $dx = \cos\theta d\theta$, the limits change from $[0, \frac{\sqrt{3}}{2}]$ to $[0, \frac{\pi}{3}]$: $$I = 2 \int_{0}^{\frac{\pi}{3}} \frac{\sin^2\theta}{\cos\theta} \cos\theta d\theta = 2 \int_{0}^{\frac{\pi}{3}} \sin^2\theta d\theta$$ Applying the identity $2\sin^2\theta = 1 - \cos(2\theta)$: $$I = \int_{0}^{\frac{\pi}{3}} (1 - \cos(2\theta)) d\theta = \left[ \theta - \frac{\sin(2\theta)}{2} \right]_{0}^{\frac{\pi}{3}}$$ Evaluating at the boundaries, we obtain the final result: $$\mathbf{\frac{\pi}{3} - \frac{\sqrt{3}}{4}}$$

Why did the $\cos\theta$ term cancel out during the trigonometric substitution?

That is the beauty of the substitution $x = \sin\theta$. When you differentiate, $dx = \cos\theta d\theta$. The denominator of our integrand was $\sqrt{1 - x^2}$, which becomes $\sqrt{1 - \sin^2\theta} = \sqrt{\cos^2\theta} = \cos\theta$. Because we have a $\cos\theta$ in the denominator and a $\cos\theta$ from the $dx$ term, they cancel out perfectly, leaving us with a simple $\sin^2\theta$ to integrate. It is a sign that we chose the right path!

JEE Advanced 2014

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

Q: Wait, why do we change $|x^2 - 1|$ to $1 - x^2$ in the Integrating Factor?

This is a fantastic question that touches on the domain of the function. The problem explicitly states that $x \in (-1, 1)$. In this interval, $x^2$ is always strictly less than $1$. Therefore, $x^2 - 1$ is always negative. Since the definition of the absolute value $|u|$ is $u$ if $u \ge 0$ and $-u$ if $u < 0$, we must write $|x^2 - 1| = -(x^2 - 1) = 1 - x^2$. Ignoring this sign change would lead to an undefined logarithm or a sign error that cascades through the entire solution.

Q: Why can't we just integrate the whole function $f(x)$ directly without splitting it?

You absolutely could, but you would be making your life much harder! When you see symmetric limits like $[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$, your mathematical intuition should immediately scream 'Symmetry!'. By splitting the integral into an odd part and an even part, the odd part vanishes to zero instantly. It is not just a shortcut; it is the elegant way to respect the geometric properties of the function.

Q: Why did the $\cos\theta$ term cancel out during the trigonometric substitution?

That is the beauty of the substitution $x = \sin\theta$. When you differentiate, $dx = \cos\theta d\theta$. The denominator of our integrand was $\sqrt{1 - x^2}$, which becomes $\sqrt{1 - \sin^2\theta} = \sqrt{\cos^2\theta} = \cos\theta$. Because we have a $\cos\theta$ in the denominator and a $\cos\theta$ from the $dx$ term, they cancel out perfectly, leaving us with a simple $\sin^2\theta$ to integrate. It is a sign that we chose the right path!

Select Answer:

Progressive Hints

Hint 1

The Spark: Look closely at the structure of the differential equation. It is a classic linear form:

\frac{d y}{d x} + P (x) y = Q (x)

. Your first task is to identify

P (x)

and

Q (x)

correctly. The hidden trap here is the domain

x \in (- 1, 1)

, which dictates how you handle the absolute value in your Integrating Factor.

Hint 2

The Arsenal: You need to calculate the Integrating Factor

I . F . = e^{\int P (x) d x}

. When you integrate

P (x) = \frac{x}{x ^{2} - 1}

, remember that since

x^{2} < 1

, the term

x^{2} - 1

is negative. You must handle the absolute value carefully:

∣ x^{2} - 1∣ = 1 - x^{2}

. This is the key to unlocking the simplification of the entire equation.

Hint 3

The Execution: Once you have the explicit function

f (x)

, you are asked to evaluate a definite integral over symmetric limits

[- \frac{3}{2}, \frac{3}{2}]

. Before you start calculating, pause and check if the function is even or odd. This will save you from a massive amount of tedious algebra!

Visualized Solution

Identify the Linear Form

The given differential equation is: $\frac{d y}{d x} + \frac{x}{x ^{2} - 1} y = \frac{x ^{4} + 2 x}{1 - x ^{2}}$
This is a Linear Differential Equation of the form: $\frac{d y}{d x} + P (x) y = Q (x)$
Here, $P (x) = \frac{x}{x ^{2} - 1}$ and $Q (x) = \frac{x ^{4} + 2 x}{1 - x ^{2}}$

Calculate Integrating Factor $(I . F .)$

Integrating Factor $(I . F .) = e^{\int P (x) d x}$
$\int P (x) d x = \int \frac{x}{x ^{2} - 1} d x$
Let $u = x^{2} - 1 \Rightarrow d u = 2 x d x$
$\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ x^{2} - 1∣$

Simplify the Integrating Factor

Since $x \in (- 1, 1)$ , we have $x^{2} < 1$ , so $∣ x^{2} - 1∣ = 1 - x^{2}$
Therefore, $\int P (x) d x = \frac{1}{2} ln (1 - x^{2}) = ln 1 - x^{2}$
$I . F . = e^{l n 1 - x^{2}} = 1 - x^{2}$

General Solution Setup

The general solution is: $y \cdot (I . F .) = \int Q (x) \cdot (I . F .) d x$
Substitute the values: $y 1 - x^{2} = \int (\frac{x ^{4} + 2 x}{1 - x ^{2}}) \cdot 1 - x^{2} d x$
Notice how the $1 - x^{2}$ terms cancel out perfectly!
Simplified integrand: $y 1 - x^{2} = \int (x^{4} + 2 x) d x$

Integrate the Right Hand Side

Integrate the polynomial: $\int (x^{4} + 2 x) d x = \frac{x ^{5}}{5} + x^{2} + C$
The equation becomes: $y 1 - x^{2} = \frac{x ^{5}}{5} + x^{2} + C$
This represents the family of curves for the differential equation.

Apply Initial Condition $f (0) = 0$

We are given the initial condition: $f (0) = 0$
This means when $x = 0$ , $y = 0$ .
Substitute into the equation: $0 \cdot 1 - 0 = \frac{0 ^{5}}{5} + 0^{2} + C$
$0 = 0 + 0 + C \Rightarrow C = 0$

Define the Function $f (x)$

With $C = 0$ , the equation simplifies to: $y 1 - x^{2} = \frac{x ^{5}}{5} + x^{2}$
Isolate $y$ to find the explicit function $f (x)$ :
$f (x) = \frac{\frac{x ^{5}}{5} + x ^{2}}{1 - x ^{2}}$
This is the function we need to integrate next.

Set up the Definite Integral

We need to evaluate: $I = \int_{- \frac{3}{2}}^{\frac{3}{2}} f (x) d x$
Substitute $f (x)$ : $I = \int_{- \frac{3}{2}}^{\frac{3}{2}} \frac{\frac{x ^{5}}{5} + x ^{2}}{1 - x ^{2}} d x$
Split the integral into two parts:
$I = \int_{- \frac{3}{2}}^{\frac{3}{2}} \frac{x ^{5}}{5 1 - x ^{2}} d x + \int_{- \frac{3}{2}}^{\frac{3}{2}} \frac{x ^{2}}{1 - x ^{2}} d x$

Analyze the Odd Function

Let $g (x) = \frac{x ^{5}}{5 1 - x ^{2}}$
Check for symmetry: $g (- x) = \frac{( - x ) ^{5}}{5 1 - ( - x ) ^{2}} = - \frac{x ^{5}}{5 1 - x ^{2}} = - g (x)$
Since $g (x)$ is an odd function, its integral over symmetric limits $[- a, a]$ is zero.
$\int_{- \frac{3}{2}}^{\frac{3}{2}} g (x) d x = 0$

Analyze the Even Function

Let $h (x) = \frac{x ^{2}}{1 - x ^{2}}$
Check for symmetry: $h (- x) = \frac{( - x ) ^{2}}{1 - ( - x ) ^{2}} = \frac{x ^{2}}{1 - x ^{2}} = h (x)$
Since $h (x)$ is an even function, $\int_{- a}^{a} h (x) d x = 2 \int_{0}^{a} h (x) d x$
Thus, $I = 0 + 2 \int_{0}^{\frac{3}{2}} \frac{x ^{2}}{1 - x ^{2}} d x$

Trigonometric Substitution

To evaluate $I = 2 \int_{0}^{\frac{3}{2}} \frac{x ^{2}}{1 - x ^{2}} d x$ , use substitution.
Let $x = sin θ \Rightarrow d x = cos θ d θ$
Change limits:
When $x = 0, θ = 0$
When $x = \frac{3}{2}, θ = \frac{π}{3}$

Simplify the Trigonometric Integral

Substitute into the integral: $I = 2 \int_{0}^{\frac{π}{3}} \frac{s i n ^{2} θ}{1 - s i n ^{2} θ} cos θ d θ$
Since $1 - sin^{2} θ = cos θ$ :
$I = 2 \int_{0}^{\frac{π}{3}} \frac{s i n ^{2} θ}{c o s θ} cos θ d θ$
The $cos θ$ terms cancel out: $I = 2 \int_{0}^{\frac{π}{3}} sin^{2} θ d θ$

Use Double Angle Identity

We need to integrate $2 sin^{2} θ$ .
Use the double angle identity: $2 sin^{2} θ = 1 - cos (2 θ)$
The integral becomes: $I = \int_{0}^{\frac{π}{3}} (1 - cos (2 θ)) d θ$

Final Evaluation

Integrate term by term: $I = [θ - \frac{s i n ( 2 θ )}{2}]_{0}^{\frac{π}{3}}$
Apply the upper limit ( $\frac{π}{3}$ ): $(\frac{π}{3} - \frac{s i n ( \frac{2 π}{3} )}{2})$
Apply the lower limit ( $0$ ): $(0 - 0) = 0$
Since $sin (\frac{2 π}{3}) = \frac{3}{2}$ :
$I = \frac{π}{3} - \frac{\frac{3}{2}}{2} = \frac{π}{3} - \frac{3}{4}$
Final Answer: $\frac{π}{3} - \frac{3}{4}$

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is:

\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}

This expression follows the standard form of a Linear Differential Equation:

\frac{d y}{d x} + P (x) y = Q (x)

. By inspection, we identify:

P (x) = \frac{x}{x ^{2} - 1}, Q (x) = \frac{x ^{4} + 2 x}{1 - x ^{2}}

Determining the Integrating Factor

The Integrating Factor (

I . F .

) is defined as

e^{\int P (x) d x}

. We compute the integral:

\int P (x) d x = \int \frac{x}{x ^{2} - 1} d x = \frac{1}{2} ln ∣ x^{2} - 1∣

Given the domain

x \in (- 1, 1)

, we note that

x^{2} < 1

, which implies

x^{2} - 1 < 0

. Therefore,

∣ x^{2} - 1∣ = 1 - x^{2}

. The

I . F .

becomes:

I . F . = e^{l n 1 - x^{2}} = 1 - x^{2}

Solving the Differential Equation

Multiplying the original differential equation by the

I . F .

simplifies the expression significantly. The equation becomes:

\frac{d}{d x} (y \cdot 1 - x^{2}) = x^{4} + 2 x

Integrating both sides with respect to

x

y 1 - x^{2} = \int (x^{4} + 2 x) d x = \frac{x ^{5}}{5} + x^{2} + C

Using the initial condition

f (0) = 0

, we find

C = 0

. Thus, the explicit function is:

f (x) = \frac{\frac{x ^{5}}{5} + x ^{2}}{1 - x ^{2}}

Final Calculation via Symmetry

We are tasked with evaluating the definite integral

I = \int_{- \frac{3}{2}}^{\frac{3}{2}} f (x) d x

. We split the integral into two parts:

I = \int_{- \frac{3}{2}}^{\frac{3}{2}} \frac{x ^{5}}{5 1 - x ^{2}} d x + \int_{- \frac{3}{2}}^{\frac{3}{2}} \frac{x ^{2}}{1 - x ^{2}} d x

The first term is an odd function integrated over symmetric limits, which evaluates to

0

. The second term is an even function, allowing us to simplify:

I = 2 \int_{0}^{\frac{3}{2}} \frac{x ^{2}}{1 - x ^{2}} d x

Using the substitution

x = sin θ

, where

d x = cos θ d θ

, the limits change from

[0, \frac{3}{2}]

[0, \frac{π}{3}]

I = 2 \int_{0}^{\frac{π}{3}} \frac{sin ^{2} θ}{cos θ} cos θ d θ = 2 \int_{0}^{\frac{π}{3}} sin^{2} θ d θ

Applying the identity

2 sin^{2} θ = 1 - cos (2 θ)

I = \int_{0}^{\frac{π}{3}} (1 - cos (2 θ)) d θ = [θ - \frac{sin ( 2 θ )}{2}]_{0}^{\frac{π}{3}}

Evaluating at the boundaries, we obtain the final result:

\frac{π}{3} - \frac{3}{4}

Common Traps & FAQs

Wait, why do we change

∣ x^{2} - 1∣

1 - x^{2}

in the Integrating Factor?

This is a fantastic question that touches on the domain of the function. The problem explicitly states that

x \in (- 1, 1)

. In this interval,

x^{2}

is always strictly less than

1

. Therefore,

x^{2} - 1

is always negative. Since the definition of the absolute value

∣ u ∣

u

u \geq 0

and

- u

u < 0

, we must write

∣ x^{2} - 1∣ = - (x^{2} - 1) = 1 - x^{2}

. Ignoring this sign change would lead to an undefined logarithm or a sign error that cascades through the entire solution.

Why can't we just integrate the whole function

f (x)

directly without splitting it?

You absolutely could, but you would be making your life much harder! When you see symmetric limits like

[- \frac{3}{2}, \frac{3}{2}]

, your mathematical intuition should immediately scream 'Symmetry!'. By splitting the integral into an odd part and an even part, the odd part vanishes to zero instantly. It is not just a shortcut; it is the elegant way to respect the geometric properties of the function.

Why did the

cos θ

term cancel out during the trigonometric substitution?

That is the beauty of the substitution

x = sin θ

. When you differentiate,

d x = cos θ d θ

. The denominator of our integrand was

1 - x^{2}

, which becomes

1 - sin^{2} θ = cos^{2} θ = cos θ

. Because we have a

cos θ

in the denominator and a

cos θ

from the

d x

term, they cancel out perfectly, leaving us with a simple

sin^{2} θ

to integrate. It is a sign that we chose the right path!

Animated Solution for Mathematics - Differential Equations: The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

Select Answer:

Visualized Solution

Identify the Linear Form

Calculate Integrating Factor $(I . F .)$

Simplify the Integrating Factor

General Solution Setup

Integrate the Right Hand Side

Apply Initial Condition $f (0) = 0$

Define the Function $f (x)$

Set up the Definite Integral

Analyze the Odd Function

Analyze the Even Function

Trigonometric Substitution

Simplify the Trigonometric Integral

Use Double Angle Identity

Final Evaluation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Determining the Integrating Factor

Solving the Differential Equation

Final Calculation via Symmetry

Similar Questions

Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}, x \in (- 1, 1)$ pass through the origin. Then $\int_{- \frac{3}{2}}^{\frac{3}{2}} f (x) d x$ is equal to

If $x = f (y)$ is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{t a n^{- 1} y}) \frac{d y}{d x} = 0,$ $y \in (- \frac{π}{2}, \frac{π}{2})$ with $f (0) = 1$ , then $f (\frac{1}{3})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ is equal to:

Let $y = y (x)$ be the solution curve of the differential equation $sin (2 x^{2}) lo g_{e} (tan x^{2}) d y + (4 x y - 42 x sin (x^{2} - \frac{π}{4})) d x = 0, 0 < x < \frac{π}{2}$ , which passes through the point $(\frac{π}{6}, 1)$ . Then $∣ y (\frac{π}{3}) ∣$ is equal to

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

Let $y = f (x)$ be the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{6} + 4 x}{1 - x ^{2}},$ $- 1 < x < 1$ such that $f (0) = 0$ . If $6 \int_{- 1/2}^{1/2} f (x) d x = 2 π - α$ then $α^{2}$ is equal to

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) = {1, 0, x \in [0, 1] otherwise$ . If $y (0) = 0$ , then $y (3/2)$ is :

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0, y (0) = 1$ . Then the value of $y (1)$ is :

If for the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + (tan x) y = \frac{2 + s e c x}{( 1 + 2 s e c x ) ^{2}},$ $x \in (\frac{- π}{2}, \frac{π}{2}), f (\frac{π}{3}) = \frac{3}{10} .$ then $f (\frac{π}{4})$ is equal to:

Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} + \frac{3 x ^{5} t a n ^{- 1} ( x ^{3} )}{( 1 + x ^{6} ) ^{\frac{3}{2}}} y = 2 x exp (\frac{x ^{3} - t a n ^{- 1} x ^{3}}{( 1 + x ^{6} )})$ pass through the origin. Then $y (1)$ is equal to :

Select Answer:

Visualized Solution

Identify the Linear Form

Calculate Integrating Factor .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (I.F.)

Simplify the Integrating Factor

General Solution Setup

Integrate the Right Hand Side

Apply Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)=0

Define the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Set up the Definite Integral

Analyze the Odd Function

Analyze the Even Function

Trigonometric Substitution

Simplify the Trigonometric Integral

Use Double Angle Identity

Final Evaluation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Determining the Integrating Factor

Solving the Differential Equation

Final Calculation via Symmetry

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=f(y) is the solution of the differential equation (1+y2)+(x−2etan−1y)dxdy​=0, y∈(−2π​,2π​) with f(0)=1, then f(3​1​) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dxdy​=(y+1)((y+1)ex2/2−x),0<x<2.1, with y(2)=0. Then the value of dxdy​ at x=1 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution curve of the differential equation x2dy+(y−x1​)dx=0;x>0 and y(1)=1, then y(21​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dxdy​+2y=f(x), where f(x)={1,0,​x∈[0,1]otherwise​. If y(0)=0, then y(3/2) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution curve of the differential equation (1+x2)dy+(y−tan−1x)dx=0,y(0)=1. Then the value of y(1) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for the solution curve y=f(x) of the differential equation dxdy​+(tanx)y=(1+2secx)22+secx​, x∈(2−π​,2π​),f(3π​)=103​​. then f(4π​) is equal to:

Calculate Integrating Factor $(I . F .)$

Apply Initial Condition $f (0) = 0$

Define the Function $f (x)$

If $x = f (y)$ is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{t a n^{- 1} y}) \frac{d y}{d x} = 0,$ $y \in (- \frac{π}{2}, \frac{π}{2})$ with $f (0) = 1$ , then $f (\frac{1}{3})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ is equal to:

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) = {1, 0, x \in [0, 1] otherwise$ . If $y (0) = 0$ , then $y (3/2)$ is :

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0, y (0) = 1$ . Then the value of $y (1)$ is :

If for the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + (tan x) y = \frac{2 + s e c x}{( 1 + 2 s e c x ) ^{2}},$ $x \in (\frac{- π}{2}, \frac{π}{2}), f (\frac{π}{3}) = \frac{3}{10} .$ then $f (\frac{π}{4})$ is equal to: