JEE Main 2022 (29 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} + (\frac{2 x ^{2} + 11 x + 13}{x ^{3} + 6 x ^{2} + 11 x + 6}) y = \frac{x + 3}{x + 1}, x > - 1$ , which passes through the point $(0, 1)$ . Then $y (1)$ is equal to:

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Look closely at the structure of the differential equation. It is a classic Linear Differential Equation of the form

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

. The complexity lies in the rational function

P (x)

. Before you panic, visualize the denominator

x^{3} + 6 x^{2} + 11 x + 6

as a product of three simple linear factors. This is the key to unlocking the entire problem.

Hint 2

Hint 2 (The Arsenal): Once you have factorized the denominator, you will see that

P (x)

is a perfect candidate for Partial Fraction Decomposition. Break it down into the form

\frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}

. This will transform a daunting integral into a sum of simple logarithmic terms, which is exactly what you need for the Integrating Factor.

Hint 3

Hint 3 (The Execution): Now, construct the Integrating Factor using the formula

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

. After you integrate the partial fractions and simplify the expression using logarithmic properties, you will find that the

e

and

ln

functions cancel out beautifully. This will leave you with a manageable algebraic expression to multiply into your differential equation.

Visualized Solution

Identifying the Differential Equation Form

The given equation is a Linear Differential Equation (LDE) of the form:
$\frac{d y}{d x} + P (x) y = Q (x)$
Where $P (x) = \frac{2 x ^{2} + 11 x + 13}{x ^{3} + 6 x ^{2} + 11 x + 6}$ and $Q (x) = \frac{x + 3}{x + 1}$

Factorizing the Denominator

Factorizing the denominator of $P (x)$ :
$x^{3} + 6 x^{2} + 11 x + 6 = (x + 1) (x + 2) (x + 3)$

Partial Fraction Decomposition

Using Partial Fraction Decomposition for $P (x)$ :
$\frac{2 x ^{2} + 11 x + 13}{( x + 1 ) ( x + 2 ) ( x + 3 )} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$

Solving for Coefficients

Solving for $A, B, C$ :
$A = \frac{2 x ^{2} + 11 x + 13}{( x + 2 ) ( x + 3 )}_{x = - 1} = \frac{2 - 11 + 13}{( 1 ) ( 2 )} = 2$
$B = \frac{2 x ^{2} + 11 x + 13}{( x + 1 ) ( x + 3 )}_{x = - 2} = \frac{8 - 22 + 13}{( - 1 ) ( 1 )} = 1$
$C = \frac{2 x ^{2} + 11 x + 13}{( x + 1 ) ( x + 2 )}_{x = - 3} = \frac{18 - 33 + 13}{( - 2 ) ( - 1 )} = - 1$

Integrating $P (x)$

Integrating $P (x)$ :
$\int P (x) d x = \int (\frac{2}{x + 1} + \frac{1}{x + 2} - \frac{1}{x + 3}) d x$
$= 2 ln (x + 1) + ln (x + 2) - ln (x + 3)$
$= ln (\frac{( x + 1 ) ^{2} ( x + 2 )}{x + 3})$

Calculating the Integrating Factor

The Integrating Factor (I.F.) is:
$I . F . = e^{\int P (x) d x} = e^{l n (\frac{( x + 1 ) ^{2} ( x + 2 )}{x + 3})}$
$I . F . = \frac{( x + 1 ) ^{2} ( x + 2 )}{x + 3}$

Setting up the General Solution

The general solution is given by:
$y \cdot (I . F .) = \int Q (x) \cdot (I . F .) d x + C$
$y \frac{( x + 1 ) ^{2} ( x + 2 )}{x + 3} = \int (\frac{x + 3}{x + 1}) (\frac{( x + 1 ) ^{2} ( x + 2 )}{x + 3}) d x + C$

Simplifying and Integrating

Simplifying the integrand:
$= \int (x + 1) (x + 2) d x + C = \int (x^{2} + 3 x + 2) d x + C$
$= \frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} + 2 x + C$

Finding the Constant $C$

Using the initial condition $(0, 1)$ :
$1 \cdot \frac{( 0 + 1 ) ^{2} ( 0 + 2 )}{0 + 3} = \frac{0 ^{3}}{3} + \frac{3 ( 0 ) ^{2}}{2} + 2 (0) + C$
$\frac{2}{3} = C ⟹ C = \frac{2}{3}$

Calculating $y (1)$

Substitute $x = 1$ and $C = \frac{2}{3}$ :
$y (1) \cdot \frac{( 1 + 1 ) ^{2} ( 1 + 2 )}{1 + 3} = \frac{1 ^{3}}{3} + \frac{3 ( 1 ) ^{2}}{2} + 2 (1) + \frac{2}{3}$
$y (1) \cdot \frac{4 \cdot 3}{4} = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3}$
$3 y (1) = 1 + \frac{3}{2} + 2 = \frac{9}{2}$

Final Result

Final calculation:
$y (1) = \frac{9}{2 \cdot 3} = \frac{3}{2}$
The correct option is (2).

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The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The given differential equation is:

\frac{d y}{d x} + (\frac{2 x ^{2} + 11 x + 13}{x ^{3} + 6 x ^{2} + 11 x + 6}) y = \frac{x + 3}{x + 1}

This is a classic Linear Differential Equation (LDE) of the form

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

. Identifying this structure is the first step toward applying the method of the Integrating Factor (I.F.).

The Algebra of Factorization

Our primary challenge is to simplify

P (x) = \frac{2 x ^{2} + 11 x + 13}{x ^{3} + 6 x ^{2} + 11 x + 6}

. The cubic denominator

x^{3} + 6 x^{2} + 11 x + 6

can be factored by testing small integer roots.

Since

x = - 1

results in zero,

(x + 1)

is a factor. Dividing the cubic by

(x + 1)

yields the quadratic

(x^{2} + 5 x + 6)

, which further factors into

(x + 2) (x + 3)

Thus, the denominator simplifies to

(x + 1) (x + 2) (x + 3)

. This transformation reduces the complex expression into a manageable form for further analysis.

The Art of Partial Fractions

We now apply Partial Fraction Decomposition to

P (x)

\frac{2 x ^{2} + 11 x + 13}{( x + 1 ) ( x + 2 ) ( x + 3 )} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}

Using the cover-up method, we determine the constants:

A = 2

B = 1

, and

C = - 1

. The expression for

P (x)

becomes:

\frac{2}{x + 1} + \frac{1}{x + 2} - \frac{1}{x + 3}

The Magic of the Integrating Factor

The Integrating Factor is defined as

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

. Integrating our partial fractions, we obtain:

\int P (x) d x = 2 ln (x + 1) + ln (x + 2) - ln (x + 3)

Using logarithmic properties, this simplifies to

ln (\frac{( x + 1 ) ^{2} ( x + 2 )}{x + 3})

. Consequently, the Integrating Factor is:

I . F . = \frac{( x + 1 ) ^{2} ( x + 2 )}{x + 3}

The Final Integration

The general solution is given by

y \cdot (I . F .) = \int Q (x) \cdot (I . F .) d x + C

. Substituting our known values:

y \cdot \frac{( x + 1 ) ^{2} ( x + 2 )}{x + 3} = \int (\frac{x + 3}{x + 1}) (\frac{( x + 1 ) ^{2} ( x + 2 )}{x + 3}) d x + C

After canceling the common terms

(x + 3)

and one

(x + 1)

, the integral simplifies significantly:

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

Performing the integration, we get:

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

The Final Result

Using the initial condition

(0, 1)

, we substitute

x = 0

and

y = 1

to find

C

1 \cdot \frac{( 1 ) ^{2} ( 2 )}{3} = 0 + 0 + 0 + C ⟹ C = \frac{2}{3}

To find

y (1)

, we substitute

x = 1

and

C = \frac{2}{3}

into the general solution:

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

3 y (1) = \frac{9}{2} ⟹ y (1) = \frac{3}{2}

Common Traps & FAQs

Wait, why can't we just separate the variables here? It looks like it might work.

That is a very common temptation! However, this is a Linear Differential Equation, not a separable one. The term

P (x) y

links

x

and

y

in a way that prevents simple separation. If you try to move all

y

terms to one side and

x

terms to the other, you will find it impossible. Always check the standard form

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

first; it is the most reliable path for these types of problems.

How do I know when to use Partial Fraction Decomposition?

Great question! You should look for it whenever you have a rational function (a polynomial divided by another polynomial) where the denominator can be easily factorized into linear or quadratic factors. In this problem, the cubic denominator

x^{3} + 6 x^{2} + 11 x + 6

is a huge hint. Whenever you see a complex rational function in a differential equation, your first instinct should be to factorize the denominator and check if partial fractions can simplify the integration.

Is there a shortcut for the Integrating Factor? This seems like a lot of algebra.

While there is no 'shortcut' that bypasses the math, there is a 'pattern' to recognize. Notice how the factors in the denominator of

P (x)

were

(x + 1), (x + 2),

and

(x + 3)

. When you integrate

\frac{1}{x + a}

, you get

ln (x + a)

. The coefficients

A, B,

and

C

you found were

2, 1,

and

- 1

. This means the integral of

P (x)

will always be a sum of logarithms, which, when exponentiated, turns into a product of powers. The 'shortcut' is simply recognizing that the structure of the I.F. is designed to cancel out the

Q (x)

term, making the final integration much simpler than it first appears.

Select Answer:

Visualized Solution

Identifying the Differential Equation Form

Factorizing the Denominator

Partial Fraction Decomposition

Solving for Coefficients

Integrating $P (x)$

Calculating the Integrating Factor

Setting up the General Solution

Simplifying and Integrating

Finding the Constant $C$

Calculating $y (1)$

Final Result

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Algebra of Factorization

The Art of Partial Fractions

The Magic of the Integrating Factor

The Final Integration

The Final Result

Similar Questions

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 :

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 curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ passing through the point $(1, 0)$ . Then $y (2)$ 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 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 $y = f (x)$ is the solution of the differential equation $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to:

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 :

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + lo g_{e} x)), x > 0, y (1) = 3$ . Then $\frac{y ^{2} ( x )}{9}$ is equal to :

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

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

Select Answer:

Visualized Solution

Identifying the Differential Equation Form

Factorizing the Denominator

Partial Fraction Decomposition

Solving for Coefficients

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

Calculating the Integrating Factor

Setting up the General Solution

Simplifying and Integrating

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

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

Final Result

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Algebra of Factorization

The Art of Partial Fractions

The Magic of the Integrating Factor

The Final Integration

The Final Result

Similar Questions

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.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

Integrating $P (x)$

Finding the Constant $C$

Calculating $y (1)$

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 curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ passing through the point $(1, 0)$ . Then $y (2)$ 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 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 $y = f (x)$ is the solution of the differential equation $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to:

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 :

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + lo g_{e} x)), x > 0, y (1) = 3$ . Then $\frac{y ^{2} ( x )}{9}$ is equal to :

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

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