JEE Main 2021 (01 September Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: 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 :

Q: Wait, why do we divide by $x^2 dx$ instead of just $dx$?

Great question! We divide by $x^2 dx$ specifically to isolate the derivative term $\frac{dy}{dx}$. In the standard form of a linear differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$, the coefficient of $\frac{dy}{dx}$ must be $1$. By dividing by $x^2 dx$, we force the equation into this standard form, allowing us to identify $P(x)$ and $Q(x)$ correctly.

Q: Why do we need the constant $C$ if we have the initial condition $y(1)=1$?

The constant $C$ is the 'memory' of the differential equation. The general solution represents an entire family of curves. The initial condition $y(1)=1$ acts as a filter, selecting the one specific curve from that family that passes through the point $(1, 1)$. Without $C$, we would have a general solution, but we wouldn't be able to find the specific value of $y$ at $x = \frac{1}{2}$.

Select Answer:

Visualized Solution

Rearranging to Standard Form

Given: $x^{2} d y + (y - \frac{1}{x}) d x = 0$
Divide by $x^{2} d x$ : $\frac{d y}{d x} + \frac{y - \frac{1}{x}}{x ^{2}} = 0$
Rearrange: $\frac{d y}{d x} + \frac{1}{x ^{2}} y = \frac{1}{x ^{3}}$

Identifying the Linear Differential Equation

Standard form: $\frac{d y}{d x} + P (x) y = Q (x)$
Compare to get $P (x) = \frac{1}{x ^{2}}$
Compare to get $Q (x) = \frac{1}{x ^{3}}$

Calculating the Integrating Factor

Formula: $I F = e^{\int P (x) d x}$
Substitute $P (x)$ : $I F = e^{\int \frac{1}{x ^{2}} d x}$
Integrate: $\int x^{- 2} d x = - \frac{1}{x}$
Result: $I F = e^{- \frac{1}{x}}$

Setting up the General Solution

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

Substitution for Integration

Let $t = - \frac{1}{x}$
Differentiate: $d t = \frac{1}{x ^{2}} d x$
Notice that $\frac{1}{x ^{3}} d x = (\frac{1}{x}) (\frac{1}{x ^{2}}) d x = (- t) d t$

Transforming the Integral

Original: $\int (\frac{1}{x}) (\frac{1}{x ^{2}}) e^{- \frac{1}{x}} d x$
Substitute $t$ and $d t$ :
$\int (- t) e^{t} d t = - \int t e^{t} d t$

Applying Integration by Parts

Use Integration by Parts on $\int t e^{t} d t$
Let $u = t \Rightarrow d u = d t$
Let $d v = e^{t} d t \Rightarrow v = e^{t}$
$\int t e^{t} d t = t e^{t} - \int e^{t} d t = t e^{t} - e^{t}$

Simplifying and Back-substitution

Integral result: $- (t e^{t} - e^{t}) = e^{t} (1 - t)$
Substitute $t = - \frac{1}{x}$ back:
$y \cdot e^{- \frac{1}{x}} = e^{- \frac{1}{x}} (1 - (- \frac{1}{x})) + C$
$y \cdot e^{- \frac{1}{x}} = e^{- \frac{1}{x}} (1 + \frac{1}{x}) + C$

Using the Initial Condition

Given condition: $y (1) = 1$
Substitute $x = 1$ and $y = 1$ :
$1 \cdot e^{- 1} = e^{- 1} (1 + 1) + C$
$e^{- 1} = 2 e^{- 1} + C$

Solving for the Constant $C$

Equation: $e^{- 1} = 2 e^{- 1} + C$
Rearrange to solve for $C$ :
$C = e^{- 1} - 2 e^{- 1}$
$C = - e^{- 1}$

The Particular Solution

Substitute $C = - e^{- 1}$ into the general solution:
$y \cdot e^{- \frac{1}{x}} = e^{- \frac{1}{x}} (1 + \frac{1}{x}) - e^{- 1}$

Finding $y (\frac{1}{2})$

Goal: Find $y$ when $x = \frac{1}{2}$
Substitute $x = \frac{1}{2}$ :
$y \cdot e^{- \frac{1}{\frac{1}{2}}} = e^{- \frac{1}{\frac{1}{2}}} (1 + \frac{1}{\frac{1}{2}}) - e^{- 1}$
Simplify exponents: $y \cdot e^{- 2} = e^{- 2} (1 + 2) - e^{- 1}$

Final Calculation Step

Equation: $y \cdot e^{- 2} = 3 e^{- 2} - e^{- 1}$
Multiply entire equation by $e^{2}$ :
$y = 3 e^{- 2} \cdot e^{2} - e^{- 1} \cdot e^{2}$
$y = 3 - e^{1}$

Conclusion and Final Answer

Final result: $y (\frac{1}{2}) = 3 - e$
Correct Option: $3 - e$
Key Takeaways:
1. Identified as a Linear Differential Equation.
2. Used $I F = e^{\int P d x}$ to simplify.
3. Solved integral using substitution and parts.

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The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The given differential equation is

x^{2} d y + (y - \frac{1}{x}) d x = 0

. While it appears intimidating, we can resolve it by transforming it into the standard form of a Linear Differential Equation.

To begin, we divide the entire equation by

x^{2} d x

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

This reveals our components:

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

and

Q (x) = \frac{1}{x ^{3}}

. We have successfully oriented the problem into the standard form

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

The Integrating Factor

To solve this, we calculate the Integrating Factor ( $I F$ ), which acts as a bridge to simplify the differential expression. The formula is defined as:

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

Substituting our

P (x) = x^{- 2}

, we compute the integral:

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

This multiplier is the key to collapsing the left-hand side of our differential equation into a single derivative.

Solving the Integral

With the

I F

determined, the general solution is given by

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

. Substituting our values, we obtain:

y \cdot e^{- \frac{1}{x}} = \int \frac{1}{x ^{3}} e^{- \frac{1}{x}} d x

To solve the integral on the right, we use the substitution

t = - \frac{1}{x}

, which implies

d t = \frac{1}{x ^{2}} d x

. Noting that

\frac{1}{x ^{3}} d x = - t d t

, the integral transforms into:

- \int t e^{t} d t

Applying Integration by Parts (

\int u d v = uv - \int v d u

) with

u = t

and

d v = e^{t} d t

, we get

- (t e^{t} - e^{t}) = e^{t} (1 - t)

. Substituting back

t = - \frac{1}{x}

, the general solution is:

y \cdot e^{- \frac{1}{x}} = e^{- \frac{1}{x}} (1 + \frac{1}{x}) + C

Applying Initial Conditions

We use the condition

y (1) = 1

to determine the constant

C

. Substituting

x = 1

and

y = 1

1 \cdot e^{- 1} = e^{- 1} (1 + 1) + C

e^{- 1} = 2 e^{- 1} + C ⟹ C = - e^{- 1}

Now, we substitute

x = \frac{1}{2}

into the specific solution to find the final value:

y \cdot e^{- 2} = e^{- 2} (1 + \frac{1}{1/2}) - e^{- 1}

y \cdot e^{- 2} = e^{- 2} (3) - e^{- 1}

Multiplying both sides by

e^{2}

, we arrive at the final result:

$y = 3 - e$

Common Traps & FAQs

Wait, why do we divide by

x^{2} d x

instead of just

d x

Great question! We divide by

x^{2} d x

specifically to isolate the derivative term

\frac{d y}{d x}

. In the standard form of a linear differential equation,

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

, the coefficient of

\frac{d y}{d x}

must be

1

. By dividing by

x^{2} d x

, we force the equation into this standard form, allowing us to identify

P (x)

and

Q (x)

correctly.

The integral

\int \frac{1}{x ^{3}} e^{- \frac{1}{x}} d x

looks impossible. How do I know to use substitution?

It is all about pattern recognition! Whenever you see a function like

e^{f (x)}

multiplied by another term, look at the derivative of the exponent. Here, the exponent is

- \frac{1}{x}

, and its derivative is

\frac{1}{x ^{2}}

. Since we have a

\frac{1}{x ^{3}}

term, we can split it into

\frac{1}{x} \cdot \frac{1}{x ^{2}}

. This

\frac{1}{x ^{2}}

is exactly what we need for the

d t

part of our substitution. Always look for the derivative of the exponent hidden within the integral!

Why do we need the constant

C

if we have the initial condition

y (1) = 1

The constant

C

is the 'memory' of the differential equation. The general solution represents an entire family of curves. The initial condition

y (1) = 1

acts as a filter, selecting the one specific curve from that family that passes through the point

(1, 1)

. Without

C

, we would have a general solution, but we wouldn't be able to find the specific value of

y

x = \frac{1}{2}

Animated Solution for Mathematics - Differential Equations: 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 :

Select Answer:

Visualized Solution

Rearranging to Standard Form

Identifying the Linear Differential Equation

Calculating the Integrating Factor

Setting up the General Solution

Substitution for Integration

Transforming the Integral

Applying Integration by Parts

Simplifying and Back-substitution

Using the Initial Condition

Solving for the Constant $C$

The Particular Solution

Finding $y (\frac{1}{2})$

Final Calculation Step

Conclusion and Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Integrating Factor

Solving the Integral

Applying Initial Conditions

Similar Questions

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 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 :

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 $x^{3} d y + (x y - 1) d x = 0, x > 0, y (\frac{1}{2}) = 3 - e$ . Then $y (1)$ is equal to

Let $y (x)$ be the solution of the differential equation $2 x^{2} d y + (e^{y} - 2 x) d x = 0, x > 0$ . If $y (e) = 1$ , then $y (1)$ 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 $y = f (x)$ 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 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 :

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 :

If $y = y (x)$ is the solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2}$ satisfying $y (1) = 1$ , then $y (1/2)$ is equal to :

Select Answer:

Visualized Solution

Rearranging to Standard Form

Identifying the Linear Differential Equation

Calculating the Integrating Factor

Setting up the General Solution

Substitution for Integration

Transforming the Integral

Applying Integration by Parts

Simplifying and Back-substitution

Using the Initial Condition

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

The Particular Solution

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(21​)

Final Calculation Step

Conclusion and Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Integrating Factor

Solving the Integral

Applying Initial Conditions

Similar Questions

.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 x(x2+ex)dy+(ex(x−2)y−x3)dx=0,x>0 passing through the point (1,0). Then y(2) 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 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; } 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 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; } Let y(x) be the solution of the differential equation 2x2dy+(ey−2x)dx=0,x>0. If y(e)=1, then y(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; } Let y=f(x) is the solution of the differential equation ey(dxdy​−1)=ex such that y(0)=0, then y(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​+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; } 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; } If y=y(x) is the solution of the differential equation xdxdy​+2y=x2 satisfying y(1)=1, then y(1/2) is equal to :

Solving for the Constant $C$

Finding $y (\frac{1}{2})$

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 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 :

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 $x^{3} d y + (x y - 1) d x = 0, x > 0, y (\frac{1}{2}) = 3 - e$ . Then $y (1)$ is equal to

Let $y (x)$ be the solution of the differential equation $2 x^{2} d y + (e^{y} - 2 x) d x = 0, x > 0$ . If $y (e) = 1$ , then $y (1)$ 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 $y = f (x)$ 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 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 :

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 :

If $y = y (x)$ is the solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2}$ satisfying $y (1) = 1$ , then $y (1/2)$ is equal to :