JEE Main 2021 (18 March Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: 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:

Q: Why did we choose $Y = y+1$ instead of just $y$?

That is an excellent observation. In differential equations, we are essentially pattern hunters. The term $(y+1)$ is the 'soul' of this equation; it dictates the behavior of the entire expression. By substituting $Y = y+1$, we collapse the complexity into a single variable, making the Bernoulli structure immediately visible. If we had kept $y$, the algebra would have become a nightmare of nested terms.

Q: Why do we use $k = -1/Y$ instead of just $1/Y$?

It is all about the derivative. If we set $k = 1/Y$, then $\frac{dk}{dx} = -\frac{1}{Y^2} \frac{dY}{dx}$. By using the negative sign in our substitution $k = -1/Y$, we ensure that the derivative term $\frac{dk}{dx}$ matches the structure of our equation perfectly. This small choice saves us from potential sign errors that often plague students in the heat of an exam.

Q: Can we solve this without the Integrating Factor?

While there are other methods for specific types of equations, the Integrating Factor method is the gold standard for linear differential equations. It is robust, systematic, and less prone to errors. In the JEE Advanced, you want a method that is reliable and repeatable, and the Integrating Factor provides exactly that.

Select Answer:

Progressive Hints

Hint 1

The Spark: Look closely at the structure of the differential equation. The term

(y + 1)

appears repeatedly, acting as a siren song for a substitution. Do not let the complexity overwhelm you; simplify the landscape by setting

Y = y + 1

. This transformation is the key to unlocking the hidden structure of the problem.

Hint 2

The Arsenal: Once you have substituted

Y = y + 1

, you will arrive at an equation of the form

\frac{d Y}{d x} + P (x) Y = Q (x) Y^{n}

. This is the classic Bernoulli differential equation. Your goal is to linearize it by dividing the entire equation by the power of

Y

that is causing the non-linearity.

Hint 3

The Execution: To linearize the Bernoulli equation, define a new variable

k = - \frac{1}{Y}

. Differentiate this with respect to

x

to find

\frac{d k}{d x}

. Substitute this into your equation to obtain a standard linear differential equation of the form

\frac{d k}{d x} + P k = Q

, which you can solve using an Integrating Factor.

Visualized Solution

Substitution: $y + 1 = Y$

Let $y + 1 = Y$
Differentiating both sides with respect to $x$ :
$\frac{d y}{d x} = \frac{d Y}{d x}$

Identifying Bernoulli's Form

Substitute $Y$ back into the original equation:
$\frac{d Y}{d x} = Y (Y e^{\frac{x ^{2}}{2}} - x)$
Expanding the terms:
$\frac{d Y}{d x} = Y^{2} e^{\frac{x ^{2}}{2}} - x Y$
Rearranging into Bernoulli's form:
$\frac{d Y}{d x} + x Y = Y^{2} e^{\frac{x ^{2}}{2}}$

Linearization via Substitution

Divide the equation by $Y^{2}$ :
$\frac{1}{Y ^{2}} \frac{d Y}{d x} + \frac{x}{Y} = e^{\frac{x ^{2}}{2}}$
Let $k = - \frac{1}{Y}$
Differentiating $k$ with respect to $x$ :
$\frac{d k}{d x} = \frac{1}{Y ^{2}} \frac{d Y}{d x}$

The Linear Differential Equation

Substitute $k$ and $\frac{d k}{d x}$ into the equation:
$\frac{d k}{d x} - x k = e^{\frac{x ^{2}}{2}}$
This is a linear differential equation of the form $\frac{d k}{d x} + P k = Q$ , where:
$P = - x$ and $Q = e^{\frac{x ^{2}}{2}}$

Finding the Integrating Factor

Integrating Factor (I.F.) $= e^{\int P d x}$
I.F. $= e^{\int - x d x}$
I.F. $= e^{- \frac{x ^{2}}{2}}$

Solving the Linear Equation

The solution is given by $k \cdot (I . F .) = \int Q \cdot (I . F .) d x$
$k e^{- \frac{x ^{2}}{2}} = \int e^{\frac{x ^{2}}{2}} \cdot e^{- \frac{x ^{2}}{2}} d x$
$k e^{- \frac{x ^{2}}{2}} = \int 1 d x$
$k e^{- \frac{x ^{2}}{2}} = x + C$

Applying Boundary Conditions

Substitute $k = - \frac{1}{y + 1}$ :
$- \frac{1}{y + 1} e^{- \frac{x ^{2}}{2}} = x + C$
Given $y (2) = 0 \Rightarrow$ at $x = 2, y = 0$ :
$- \frac{1}{0 + 1} e^{- \frac{2 ^{2}}{2}} = 2 + C$
$- e^{- 2} = 2 + C \Rightarrow C = - 2 - e^{- 2}$

Finding $y + 1$ at $x = 1$

At $x = 1$ :
$- \frac{1}{y + 1} e^{- \frac{1}{2}} = 1 + (- 2 - e^{- 2})$
$\frac{1}{y + 1} e^{- \frac{1}{2}} = 1 + e^{- 2}$
$\frac{1}{y + 1} = e^{\frac{1}{2}} (1 + \frac{1}{e ^{2}}) = e^{\frac{1}{2}} (\frac{e ^{2} + 1}{e ^{2}})$
$\frac{1}{y + 1} = \frac{e ^{2} + 1}{e ^{\frac{3}{2}}} \Rightarrow y + 1 = \frac{e ^{\frac{3}{2}}}{e ^{2} + 1}$

Final Calculation for $\frac{d y}{d x}$

Original equation: $\frac{d y}{d x} = (y + 1)^{2} e^{\frac{x ^{2}}{2}} - x (y + 1)$
At $x = 1$ :
$\frac{d y}{d x} = (\frac{e ^{\frac{3}{2}}}{e ^{2} + 1})^{2} e^{\frac{1}{2}} - 1 (\frac{e ^{\frac{3}{2}}}{e ^{2} + 1})$
$\frac{d y}{d x} = \frac{e ^{3} \cdot e ^{\frac{1}{2}}}{( e ^{2} + 1 ) ^{2}} - \frac{e ^{\frac{3}{2}}}{e ^{2} + 1}$
$\frac{d y}{d x} = \frac{e ^{\frac{7}{2}} - e ^{\frac{3}{2}} ( e ^{2} + 1 )}{( e ^{2} + 1 ) ^{2}} = \frac{e ^{\frac{7}{2}} - e ^{\frac{7}{2}} - e ^{\frac{3}{2}}}{( e ^{2} + 1 ) ^{2}}$
$\frac{d y}{d x} = - \frac{e ^{\frac{3}{2}}}{( e ^{2} + 1 ) ^{2}}$

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The given differential equation is:

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

Notice the repetition of the term

(y + 1)

. Whenever you see a repeating cluster in a differential equation, your intuition should scream substitution. Let us define a new variable

Y = y + 1

Differentiating with respect to

x

, we get

\frac{d y}{d x} = \frac{d Y}{d x}

. Substituting this into the original equation yields:

\frac{d Y}{d x} = Y (Y e^{x^{2} /2} - x)

Expanding this, we arrive at:

\frac{d Y}{d x} = Y^{2} e^{x^{2} /2} - x Y

Linearizing the Bernoulli Equation

Rearranging the terms, we obtain:

\frac{d Y}{d x} + x Y = Y^{2} e^{x^{2} /2}

This is a classic Bernoulli differential equation. To linearize it, we divide the entire equation by

Y^{2}

\frac{1}{Y ^{2}} \frac{d Y}{d x} + \frac{x}{Y} = e^{x^{2} /2}

Now, we introduce the substitution

k = - \frac{1}{Y}

. Differentiating

k

with respect to

x

gives:

\frac{d k}{d x} = \frac{1}{Y ^{2}} \frac{d Y}{d x}

Substituting

k

and

\frac{d k}{d x}

into our equation, we transform it into a standard linear differential equation:

\frac{d k}{d x} - x k = e^{x^{2} /2}

Solving the Linear Equation

This equation is of the form

\frac{d k}{d x} + P k = Q

, where

P = - x

and

Q = e^{x^{2} /2}

. The Integrating Factor (I.F.) is:

I . F . = e^{\int P d x} = e^{\int - x d x} = e^{- x^{2} /2}

Multiplying the linear equation by the I.F., we get:

k e^{- x^{2} /2} = \int e^{x^{2} /2} \cdot e^{- x^{2} /2} d x

The exponentials cancel out to become

1

, simplifying the integral to:

k e^{- x^{2} /2} = \int 1 d x = x + C

Applying Boundary Conditions

We are given the boundary condition

y (2) = 0

. Since

k = - \frac{1}{y + 1}

, at

x = 2

we have:

k = - \frac{1}{0 + 1} = - 1

Plugging these values into our solution:

- 1 \cdot e^{- 2^{2} /2} = 2 + C \Rightarrow - e^{- 2} = 2 + C

Thus, the constant is

C = - 2 - e^{- 2}

Final Calculation

To find the value of

y + 1

x = 1

, we use the relation

k = (x + C) e^{x^{2} /2}

. Substituting

x = 1

and our value of

C

k = (1 - 2 - e^{- 2}) e^{1/2} = (- 1 - e^{- 2}) e^{1/2}

Since

k = - \frac{1}{y + 1}

, we have

y + 1 = - \frac{1}{k}

. Therefore:

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

This simplifies to the final result:

$y + 1 = \frac{e ^{1/2}}{1 + e ^{- 2}}$

Common Traps & FAQs

Why did we choose

Y = y + 1

instead of just

y

That is an excellent observation. In differential equations, we are essentially pattern hunters. The term

(y + 1)

is the 'soul' of this equation; it dictates the behavior of the entire expression. By substituting

Y = y + 1

, we collapse the complexity into a single variable, making the Bernoulli structure immediately visible. If we had kept

y

, the algebra would have become a nightmare of nested terms.

Why do we use

k = - 1/ Y

instead of just

1/ Y

It is all about the derivative. If we set

k = 1/ Y

, then

\frac{d k}{d x} = - \frac{1}{Y ^{2}} \frac{d Y}{d x}

. By using the negative sign in our substitution

k = - 1/ Y

, we ensure that the derivative term

\frac{d k}{d x}

matches the structure of our equation perfectly. This small choice saves us from potential sign errors that often plague students in the heat of an exam.

Can we solve this without the Integrating Factor?

While there are other methods for specific types of equations, the Integrating Factor method is the gold standard for linear differential equations. It is robust, systematic, and less prone to errors. In the JEE Advanced, you want a method that is reliable and repeatable, and the Integrating Factor provides exactly that.

Animated Solution for Mathematics - Differential Equations: 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:

Select Answer:

Visualized Solution

Substitution: $y + 1 = Y$

Identifying Bernoulli's Form

Linearization via Substitution

The Linear Differential Equation

Finding the Integrating Factor

Solving the Linear Equation

Applying Boundary Conditions

Finding $y + 1$ at $x = 1$

Final Calculation for $\frac{d y}{d x}$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Linearizing the Bernoulli Equation

Solving the Linear Equation

Applying Boundary Conditions

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 $y = f (x)$ 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 :

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 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 $(1 + x^{2}) \frac{d y}{d x} + y = e^{t a n^{- 1} x}, y (1) = 0$ . Then $y (0)$ 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 :

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to

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:

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

Substitution: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y+1=Y

Identifying Bernoulli's Form

Linearization via Substitution

The Linear Differential Equation

Finding the Integrating Factor

Solving the Linear Equation

Applying Boundary Conditions

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

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Linearizing the Bernoulli Equation

Solving the Linear Equation

Applying Boundary Conditions

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=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; } 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 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 of the differential equation (1+x2)dxdy​+y=etan−1x,y(1)=0. Then y(0) 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; } Let y=y(x) be the solution of the differential equation xloge​xdxdy​+y=x2loge​x,(x>1). If y(2)=2, then y(e) 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 cosec2xdy+2dx=(1+ycos2x)cosec2xdx, with y(4π​)=0. Then, the value of (y(0)+1)2 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function y=f(x) is the solution of the differential equation dxdy​+x2−1xy​=1−x2​x4+2x​ in (−1,1) satisfying f(0)=0. Then ∫−3​/23​/2​f(x)dx is

Substitution: $y + 1 = Y$

Finding $y + 1$ at $x = 1$

Final Calculation for $\frac{d y}{d x}$

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 $y = f (x)$ 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 :

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 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 $(1 + x^{2}) \frac{d y}{d x} + y = e^{t a n^{- 1} x}, y (1) = 0$ . Then $y (0)$ 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 :

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to

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:

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