How do I know the Integrating Factor is $\sqrt{1-x^2}$?

The Integrating Factor is defined as $e^{\int P(x) dx}$. Here, $P(x) = -\frac{x}{1-x^2}$. If we let $u = 1-x^2$, then $du = -2x dx$, so $-x dx = \frac{1}{2} du$. The integral becomes $\int \frac{1}{2u} du = \frac{1}{2} \ln(u) = \ln(\sqrt{u})$. When you raise $e$ to the power of $\ln(\sqrt{1-x^2})$, the $e$ and $\ln$ functions cancel each other out, leaving you with $\sqrt{1-x^2}$.

JEE Main 2024 (30 Jan Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) 3 (1 - x^{2})] d x, - 1 < x < 1, y (0) = 0$ . If $y (\frac{1}{2}) = \frac{m}{n}$ , $m$ and $n$ are coprime numbers, then $m + n$ is equal to_____,

Q: Wait, why do we divide by $(1-x^2)$? Isn't that risky if $x=1$ or $x=-1$?

That is a very sharp observation! In many differential equations, dividing by a term that could be zero is dangerous. However, the problem explicitly states that $-1 < x < 1$. In this open interval, $1-x^2$ is strictly positive and never zero, so our division is mathematically sound and necessary to reach the standard linear form.

Q: Why is the constant of integration $C$ equal to zero?

We determine $C$ using the initial condition $y(0) = 0$. When we substitute $x=0$ and $y=0$ into our general solution $y \sqrt{1-x^2} = \sqrt{3}(\frac{x^4}{4} + 2x) + C$, the left side becomes $0 \cdot \sqrt{1-0} = 0$. On the right side, the polynomial terms also evaluate to zero. Thus, $0 = 0 + C$, which forces $C$ to be zero. It is a common feature in these problems to have the constant vanish, but always check it carefully!

Enter Numerical Value:

Progressive Hints

Hint 1

Look closely at the structure of the given differential equation. It is a first-order linear differential equation in disguise. Try dividing the entire equation by

(1 - x^{2}) d x

to isolate

\frac{d y}{d x}

and identify the functions

P (x)

and

Q (x)

in the standard form

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

Hint 2

Once you have identified

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

, the next step is to find the Integrating Factor (

I F

). Recall that

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

. Use the substitution

t = 1 - x^{2}

to evaluate the integral of

P (x)

. You will find that the

I F

simplifies beautifully to

1 - x^{2}

Hint 3

Now, multiply the entire standard form equation by your

I F

. The left side will become the derivative of the product

y \cdot I F

, and the right side will simplify significantly because the

1 - x^{2}

in the denominator of

Q (x)

will cancel out with the

I F

. The resulting equation is

y 1 - x^{2} = \int 3 (x^{3} + 2) d x

. Integrate this and use the initial condition

y (0) = 0

to find the constant

C

Visualized Solution

Standard Form of LDE

Original Equation: $(1 - x^{2}) d y = [x y + (x^{3} + 2) 3 (1 - x^{2})] d x$
Divide by $(1 - x^{2}) d x$ to isolate $\frac{d y}{d x}$ .
$\frac{d y}{d x} - \frac{x}{1 - x ^{2}} y = \frac{( x ^{3} + 2 ) 3}{1 - x ^{2}}$

Identifying $P (x)$ and $Q (x)$

Compare with standard form: $\frac{d y}{d x} + P (x) y = Q (x)$
$P (x) = - \frac{x}{1 - x ^{2}}$
$Q (x) = \frac{( x ^{3} + 2 ) 3}{1 - x ^{2}}$

The Integrating Factor $I F$

Formula for Integrating Factor: $I F = e^{\int P (x) d x}$
Substitute $P (x)$ : $I F = e^{\int - \frac{x}{1 - x ^{2}} d x}$
Let $1 - x^{2} = t ⟹ - 2 x d x = d t$

Calculating the $I F$

$\int - \frac{x}{1 - x ^{2}} d x = \frac{1}{2} \int \frac{d t}{t} = \frac{1}{2} ln (t)$
Substitute back $t = 1 - x^{2}$ : $\frac{1}{2} ln (1 - x^{2}) = ln 1 - x^{2}$
$I F = e^{l n 1 - x^{2}} = 1 - x^{2}$

General Solution Structure

Formula: $y \cdot I F = \int Q (x) \cdot I F d x$
Substitute $I F$ and $Q (x)$ :
$y 1 - x^{2} = \int \frac{( x ^{3} + 2 ) 3}{1 - x ^{2}} \cdot 1 - x^{2} d x$

Integrating the Right Side

Cancel $1 - x^{2}$ : $y 1 - x^{2} = \int 3 (x^{3} + 2) d x$
Integrate: $\int 3 (x^{3} + 2) d x = 3 (\frac{x ^{4}}{4} + 2 x) + C$
Current Equation: $y 1 - x^{2} = 3 (\frac{x ^{4}}{4} + 2 x) + C$

Applying $y (0) = 0$

Initial Condition: $y (0) = 0$ (When $x = 0, y = 0$ )
Substitute into equation: $0 \cdot 1 - 0 = 3 (\frac{0}{4} + 0) + C$
Solve for $C$ : $0 = 0 + C ⟹ C = 0$
Specific Solution: $y 1 - x^{2} = 3 (\frac{x ^{4}}{4} + 2 x)$

Evaluating at $x = \frac{1}{2}$

We need to find $y (\frac{1}{2})$ .
Substitute $x = \frac{1}{2}$ :
$y 1 - (\frac{1}{2})^{2} = 3 (\frac{( 1/2 ) ^{4}}{4} + 2 (\frac{1}{2}))$
$y 1 - \frac{1}{4} = 3 (\frac{1/16}{4} + 1)$

Simplifying the Expression

Left side: $y \frac{3}{4} = y \frac{3}{2}$
Right side: $3 (\frac{1}{64} + 1) = 3 (\frac{65}{64})$
Equation: $y \frac{3}{2} = 3 \frac{65}{64}$

Final Value of $m + n$

Cancel $3$ : $y \frac{1}{2} = \frac{65}{64}$
Multiply by 2: $y = \frac{65}{32}$
Given $y (\frac{1}{2}) = \frac{m}{n}$ , so $m = 65, n = 32$ .
Check coprime: $gcd (65, 32) = 1$ .
$m + n = 65 + 32 = 97$ .

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The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The given differential equation is:

(1 - x^{2}) d y = [x y + (x^{3} + 2) 3 (1 - x^{2})] d x

To solve this, we aim to transform it into the standard linear form:

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

Dividing the entire equation by

(1 - x^{2}) d x

, we obtain:

\frac{d y}{d x} - \frac{x}{1 - x ^{2}} y = \frac{( x ^{3} + 2 ) 3}{1 - x ^{2}}

Determining the Integrating Factor

From the standard form, we identify:

P (x) = - \frac{x}{1 - x ^{2}}, Q (x) = \frac{( x ^{3} + 2 ) 3}{1 - x ^{2}}

The Integrating Factor (

I F

) is defined as:

I F = e^{\int P (x) d x} = e^{\int - \frac{x}{1 - x ^{2}} d x}

Using the substitution

t = 1 - x^{2}

, where

d t = - 2 x d x

, the integral becomes:

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

Thus, the Integrating Factor is:

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

Solving the Differential Equation

Multiplying the standard form equation by the

I F

, the left side becomes the derivative of the product

y \cdot 1 - x^{2}

\frac{d}{d x} (y 1 - x^{2}) = 3 (x^{3} + 2)

Integrating both sides with respect to

x

y 1 - x^{2} = 3 \int (x^{3} + 2) d x

y 1 - x^{2} = 3 (\frac{x ^{4}}{4} + 2 x) + C

Applying Initial Conditions and Final Calculation

Using the initial condition

y (0) = 0

, we substitute

x = 0

and

y = 0

into the equation:

0 \cdot 1 - 0 = 3 (0 + 0) + C ⟹ C = 0

The general solution is therefore:

y 1 - x^{2} = 3 (\frac{x ^{4}}{4} + 2 x)

To find the value at

x = \frac{1}{2}

y 1 - \frac{1}{4} = 3 (\frac{( 1/2 ) ^{4}}{4} + 2 (\frac{1}{2}))

y \frac{3}{2} = 3 (\frac{1}{64} + 1) = 3 (\frac{65}{64})

Canceling

3

from both sides:

\frac{y}{2} = \frac{65}{64} ⟹ y = \frac{65}{32}

Given

y = \frac{m}{n} = \frac{65}{32}

, where

m = 65

and

n = 32

are coprime, the final result is: $m + n = 65 + 32 = 97$

Common Traps & FAQs

Wait, why do we divide by

(1 - x^{2})

? Isn't that risky if

x = 1

x = - 1

That is a very sharp observation! In many differential equations, dividing by a term that could be zero is dangerous. However, the problem explicitly states that

- 1 < x < 1

. In this open interval,

1 - x^{2}

is strictly positive and never zero, so our division is mathematically sound and necessary to reach the standard linear form.

How do I know the Integrating Factor is

1 - x^{2}

The Integrating Factor is defined as

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

. Here,

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

. If we let

u = 1 - x^{2}

, then

d u = - 2 x d x

, so

- x d x = \frac{1}{2} d u

. The integral becomes

\int \frac{1}{2 u} d u = \frac{1}{2} ln (u) = ln (u)

. When you raise

e

to the power of

ln (1 - x^{2})

, the

e

and

ln

functions cancel each other out, leaving you with

1 - x^{2}

Why is the constant of integration

C

equal to zero?

We determine

C

using the initial condition

y (0) = 0

. When we substitute

x = 0

and

y = 0

into our general solution

y 1 - x^{2} = 3 (\frac{x ^{4}}{4} + 2 x) + C

, the left side becomes

0 \cdot 1 - 0 = 0

. On the right side, the polynomial terms also evaluate to zero. Thus,

0 = 0 + C

, which forces

C

to be zero. It is a common feature in these problems to have the constant vanish, but always check it carefully!

Enter Numerical Value:

Visualized Solution

Standard Form of LDE

Identifying $P (x)$ and $Q (x)$

The Integrating Factor $I F$

Calculating the $I F$

General Solution Structure

Integrating the Right Side

Applying $y (0) = 0$

Evaluating at $x = \frac{1}{2}$

Simplifying the Expression

Final Value of $m + n$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Determining the Integrating Factor

Solving the Differential Equation

Applying Initial Conditions and Final Calculation

Similar Questions

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 $y$ be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) 1 - x^{2}] d x, - 1 < x < 1$ and $y (0) = 0$ . If $\int_{- 1/2}^{1/2} 1 - x^{2} y (x) d x = k$ , then $k^{- 1}$ is equal to :

Let $α x = exp (x^{β} y^{γ})$ be the solution of the differential equation $2 x^{2} y d y - (1 - x y^{2}) d x = 0, x > 0, y (2) = lo g_{e} 2$ . Then $α + β - γ$ equals :

Let $y = y (x)$ be the solution of the differential equation $x^{3} d y + (x y - 1) d x = 0, x > 0, y (\frac{1}{2}) = 3 - e$ . Then $y (1)$ is equal to

If for $x \geq 0$ , $y = y (x)$ is the solution of the differential equation, $(1 + x) \frac{d y}{d x} = ((1 + x)^{2} + (y - 3))$ , If $y (2) = 0$ then $y (3) = ?$

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 :

Consider the differential equation, $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If value of y is 1 when x = 1, then the value of x for which y = 2, is :

Let $y = y (x), x > 1$ , be the solution of the differential equation $(x - 1) \frac{d y}{d x} + 2 x y = \frac{1}{x - 1}$ , with $y (2) = \frac{1 + e ^{4}}{2 e ^{4}}$ . If $y (3) = \frac{e ^{α} + 1}{β e ^{α}}$ , then the value of $α + β$ is equal to _______.

The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq = 0)$ with $y (1) = 1$ , is

Let $y = y (x)$ be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{t a n^{- 1} x}, y (1) = 0$ . Then $y (0)$ is

Enter Numerical Value:

Visualized Solution

Standard Form of LDE

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(x)

The Integrating Factor .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IF

Calculating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IF

General Solution Structure

Integrating the Right Side

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

Evaluating at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=21​

Simplifying the Expression

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Determining the Integrating Factor

Solving the Differential Equation

Applying Initial Conditions and Final Calculation

Similar Questions

.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 be the solution of the differential equation (1−x2)dy=[xy+(x3+2)1−x2​]dx,−1<x<1 and y(0)=0. If ∫−1/21/2​1−x2​y(x)dx=k, then k−1 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let αx=exp(xβyγ) be the solution of the differential equation 2x2ydy−(1−xy2)dx=0,x>0,y(2)=loge​2​. Then α+β−γ equals :

.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 x3dy+(xy−1)dx=0,x>0,y(21​)=3−e. Then y(1) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for x≥0, y=y(x) is the solution of the differential equation, (1+x)dxdy​=((1+x)2+(y−3)), If y(2)=0 then y(3)=?

.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; } Consider the differential equation, y2dx+(x−y1​)dy=0. If value of y is 1 when x = 1, then the value of x for which y = 2, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the differential equation xdxdy​+2y=x2(x=0) with y(1)=1, 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 of the differential equation (1+x2)dxdy​+y=etan−1x,y(1)=0. Then y(0) is

Identifying $P (x)$ and $Q (x)$

The Integrating Factor $I F$

Calculating the $I F$

Applying $y (0) = 0$

Evaluating at $x = \frac{1}{2}$

Final Value of $m + n$

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 $y$ be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) 1 - x^{2}] d x, - 1 < x < 1$ and $y (0) = 0$ . If $\int_{- 1/2}^{1/2} 1 - x^{2} y (x) d x = k$ , then $k^{- 1}$ is equal to :

Let $α x = exp (x^{β} y^{γ})$ be the solution of the differential equation $2 x^{2} y d y - (1 - x y^{2}) d x = 0, x > 0, y (2) = lo g_{e} 2$ . Then $α + β - γ$ equals :

Let $y = y (x)$ be the solution of the differential equation $x^{3} d y + (x y - 1) d x = 0, x > 0, y (\frac{1}{2}) = 3 - e$ . Then $y (1)$ is equal to

If for $x \geq 0$ , $y = y (x)$ is the solution of the differential equation, $(1 + x) \frac{d y}{d x} = ((1 + x)^{2} + (y - 3))$ , If $y (2) = 0$ then $y (3) = ?$

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 :

Consider the differential equation, $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If value of y is 1 when x = 1, then the value of x for which y = 2, is :

The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq = 0)$ with $y (1) = 1$ , is

Let $y = y (x)$ be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{t a n^{- 1} x}, y (1) = 0$ . Then $y (0)$ is