JEE Main 2019 (8 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation, $(x^{2} + 1) \frac{d y}{d x} + 2 x (x^{2} + 1) y = 1$ such that $y (0) = 0$ . If $a y (1) = \frac{π}{32}$ , then the value of 'a' is :

Q: How do I know if I have correctly identified the standard form?

The standard form is $\frac{dy}{dx} + P(x)y = Q(x)$. If you have a coefficient in front of $\frac{dy}{dx}$ (like the $(x^2+1)$ in this problem), you must divide the entire equation by that coefficient first. If you don't, your $P(x)$ and $Q(x)$ will be wrong, and the whole solution will collapse. Always normalize the $\frac{dy}{dx}$ term to have a coefficient of 1.

Select Answer:

Visualized Solution

Standard Form of the LDE

The given differential equation is a Linear Differential Equation (LDE).
We need to bring it to the standard form: $\frac{d y}{d x} + P (x) y = Q (x)$
Rearranging the terms, we get: $\frac{d y}{d x} + \frac{2 x}{x ^{2} + 1} y = \frac{1}{( x ^{2} + 1 ) ^{2}}$

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

Comparing our equation with the standard form $\frac{d y}{d x} + P (x) y = Q (x)$
We can identify the functions $P (x)$ and $Q (x)$ .
$P (x) = \frac{2 x}{x ^{2} + 1}$
$Q (x) = \frac{1}{( x ^{2} + 1 ) ^{2}}$

Calculating the Integrating Factor ( $I . F .$ )

The Integrating Factor (I.F.) is calculated using the formula: $I . F . = e^{\int P (x) d x}$
Substituting $P (x)$ : $I . F . = e^{\int \frac{2 x}{x ^{2} + 1} d x}$
Notice that the numerator $2 x$ is the exact derivative of the denominator $x^{2} + 1$ .
Therefore, the integral is $ln (x^{2} + 1)$ .
$I . F . = e^{l n (x^{2} + 1)} = x^{2} + 1$

General Solution Equation

The general solution of an LDE is given by: $y \cdot (I . F .) = \int Q (x) \cdot (I . F .) d x + C$
Substituting the values of $I . F .$ and $Q (x)$ into this formula:
$y (x^{2} + 1) = \int \frac{1}{( x ^{2} + 1 ) ^{2}} \cdot (x^{2} + 1) d x + C$

Integrating the Right-Hand Side

Simplifying the integrand by canceling out $(x^{2} + 1)$ :
$y (x^{2} + 1) = \int \frac{1}{x ^{2} + 1} d x + C$
This is a standard integral. The integral of $\frac{1}{x ^{2} + 1}$ is $tan^{- 1} x$ .
$y (x^{2} + 1) = tan^{- 1} x + C$

Applying the Initial Condition

We are given the initial condition: $y (0) = 0$ .
This means when $x = 0$ , $y = 0$ . Let's substitute these values to find the constant $C$ .
$0 \cdot (0^{2} + 1) = tan^{- 1} (0) + C$
Since $tan^{- 1} (0) = 0$ , we get $0 = 0 + C$ , which implies $C = 0$ .

The Specific Solution $y (x)$

Substituting $C = 0$ back into our general solution equation:
$y (x^{2} + 1) = tan^{- 1} x$
Isolating $y$ to get the specific solution:
$y (x) = \frac{t a n ^{- 1} x}{x ^{2} + 1}$

Evaluating $y (1)$

The problem asks us to use the value of $y (1)$ . Let's calculate it by substituting $x = 1$ .
$y (1) = \frac{t a n ^{- 1} ( 1 )}{1 ^{2} + 1}$
We know that $tan^{- 1} (1) = \frac{π}{4}$ .
$y (1) = \frac{\frac{π}{4}}{2} = \frac{π}{8}$

Using the Given Condition

The problem provides a specific condition: $a y (1) = \frac{π}{32}$
We just found that $y (1) = \frac{π}{8}$ . Let's substitute this value.
$a \cdot \frac{π}{8} = \frac{π}{32}$

Final Value of 'a'

Now, we solve for $a$ by cross-multiplying:
$a = \frac{π}{32} \cdot \frac{8}{π} = \frac{1}{4}$
To find $a$ , we square both sides of the equation:
$a = (\frac{1}{4})^{2} = \frac{1}{16}$
Final Answer: Option (2)

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The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The given differential equation is:

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

This equation initially appears complex, but it follows the structure of a Linear Differential Equation (LDE). To solve it, we must first normalize the equation into the standard form:

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

Dividing the entire equation by

(x^{2} + 1)

, we obtain:

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

From this, we identify the components:

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

and

Q (x) = \frac{1}{( x ^{2} + 1 ) ^{2}}

The Magic of the Integrating Factor

The Integrating Factor (

I . F .

) is defined as

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

. Substituting our

P (x)

, we get:

I . F . = e^{\int \frac{2 x}{x ^{2} + 1} d x}

Since the numerator

2 x

is the derivative of the denominator

x^{2} + 1

, the integral evaluates to

ln (x^{2} + 1)

. Applying the property

e^{l n (u)} = u

, the

I . F .

simplifies elegantly:

I . F . = x^{2} + 1

The Integration

Multiplying the standard form equation by the

I . F .

, the left side becomes the derivative of the product

y \cdot (I . F .)

. We then integrate both sides:

\frac{d}{d x} [y (x^{2} + 1)] = \frac{1}{( x ^{2} + 1 ) ^{2}} \cdot (x^{2} + 1)

\frac{d}{d x} [y (x^{2} + 1)] = \frac{1}{x ^{2} + 1}

Integrating with respect to

x

yields the general solution:

y (x^{2} + 1) = tan^{- 1} x + C

Final Calculation

We are given the initial condition

y (0) = 0

. Substituting

x = 0

and

y = 0

into the general solution:

0 (0^{2} + 1) = tan^{- 1} (0) + C \Rightarrow C = 0

Thus, the specific solution is:

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

Evaluating at

x = 1

y (1) = \frac{tan ^{- 1} ( 1 )}{1 ^{2} + 1} = \frac{π /4}{2} = \frac{π}{8}

Given the condition

a \cdot y (1) = \frac{π}{32}

, we substitute our result:

a \cdot \frac{π}{8} = \frac{π}{32} \Rightarrow a = \frac{1}{4}

Squaring both sides, we find the final value:

$a = \frac{1}{16}$

Common Traps & FAQs

Wait, why can't I just separate the variables? This looks like it could be separable.

That is a great instinct to have, but look closely at the term

2 x (x^{2} + 1) y

. Because of the addition of the

y

term, you cannot isolate all

y

's on one side and all

x

's on the other. This is the hallmark of a Linear Differential Equation. When you see

y

and

d y / d x

appearing linearly, always think of the Integrating Factor method.

How do I know if I have correctly identified the standard form?

The standard form is

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

. If you have a coefficient in front of

\frac{d y}{d x}

(like the

(x^{2} + 1)

in this problem), you must divide the entire equation by that coefficient first. If you don't, your

P (x)

and

Q (x)

will be wrong, and the whole solution will collapse. Always normalize the

\frac{d y}{d x}

term to have a coefficient of 1.

Is it possible to make a mistake with the constant

C

Absolutely! The most common mistake is forgetting to include

+ C

during the integration step. Even if

C

turns out to be zero, you must write it down and solve for it using the initial condition

y (0) = 0

. Never assume

C = 0

without proving it mathematically.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation, $(x^{2} + 1) \frac{d y}{d x} + 2 x (x^{2} + 1) y = 1$ such that $y (0) = 0$ . If $a y (1) = \frac{π}{32}$ , then the value of 'a' is :

Select Answer:

Visualized Solution

Standard Form of the LDE

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

Calculating the Integrating Factor ( $I . F .$ )

General Solution Equation

Integrating the Right-Hand Side

Applying the Initial Condition

The Specific Solution $y (x)$

Evaluating $y (1)$

Using the Given Condition

Final Value of 'a'

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Integration

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Standard Form of the 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)

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

General Solution Equation

Integrating the Right-Hand Side

Applying the Initial Condition

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

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

Using the Given Condition

Final Value of 'a'

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Integration

Final Calculation

Similar Questions

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Calculating the Integrating Factor ( $I . F .$ )

The Specific Solution $y (x)$

Evaluating $y (1)$

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Let $y = y (x)$ be the solution of the differential equation $cosec^{2} x d y + 2 d x = (1 + y cos 2 x) cosec^{2} x d x$ , with $y (\frac{π}{4}) = 0$ . Then, the value of $(y (0) + 1)^{2}$ is equal to:

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