JEE Main 2024 (01 Feb Shift 2)

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Animated Solution for Mathematics - Differential Equations: If $\frac{d x}{d y} = \frac{1 + x - y ^{2}}{y}, x (1) = 1$ , then $5 x (2)$ is equal to :

Q: Why did the integrating factor $IF$ become $1/y$?

Great observation. We calculated $IF = e^{\int -\frac{1}{y} dy}$. The integral of $-1/y$ is $-\ln(y)$. So, $IF = e^{-\ln(y)}$. Using the logarithmic property $a \ln(b) = \ln(b^a)$, this becomes $e^{\ln(y^{-1})}$. Since $e^{\ln(u)} = u$, we are left with $y^{-1}$, which is $1/y$. It is a beautiful simplification!

Q: Why do we multiply the final equation by $y$ at the end?

We do that to isolate $x$. Our solution at that stage was in the form $\frac{x}{y} = \dots$. To find the explicit function $x(y)$, we simply multiply both sides by $y$. It is just the final algebraic cleanup to make the function easy to evaluate for $x(2)$.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. Look closely at the structure of the differential equation. It is not in the standard

d y / d x

form. Instead, it is given as

d x / d y

. This is a subtle hint to treat

x

as the dependent variable and

y

as the independent variable. Rearrange the equation to isolate the linear terms in

x

Hint 2

Hint 2: The Arsenal. Once you have rearranged the equation into the form

d x / d y + P (y) x = Q (y)

, you need to identify the integrating factor. Recall that for this form, the integrating factor is

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

. Be extremely careful with the negative sign in

P (y)

Hint 3

Hint 3: The Execution. After finding the integrating factor, the general solution is given by

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

. Substitute your

I F

and

Q (y)

into this formula and integrate. Don't forget the constant of integration

C

, which you will determine using the initial condition

x (1) = 1

Visualized Solution

Problem Statement

Given differential equation: $\frac{d x}{d y} = \frac{1 + x - y ^{2}}{y}$
Initial condition: $x (1) = 1$
Target: Find the value of $5 x (2)$

Converting to Standard Form

Split the fraction on the RHS: $\frac{d x}{d y} = \frac{1}{y} + \frac{x}{y} - y$
Rearrange terms: $\frac{d x}{d y} - \frac{1}{y} x = \frac{1 - y ^{2}}{y}$
This matches the standard linear form: $\frac{d x}{d y} + P (y) x = Q (y)$

Identifying $P (y)$ and $Q (y)$

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

Integrating Factor (IF)

Formula for Integrating Factor: $I F = e^{\int P (y) d y}$
Substitute $P (y)$ : $I F = e^{\int - \frac{1}{y} d y}$
Integrate: $I F = e^{- l n y}$
Simplify using log properties: $I F = y^{- 1} = \frac{1}{y}$

General Solution Setup

General solution formula: $x \cdot I F = \int Q (y) \cdot I F d y$
Substitute $I F$ and $Q (y)$ : $x \cdot \frac{1}{y} = \int (\frac{1 - y ^{2}}{y}) \cdot \frac{1}{y} d y$
Simplify the integrand: $\frac{x}{y} = \int \frac{1 - y ^{2}}{y ^{2}} d y$

Evaluating the Integral

Split the integral: $\int (\frac{1}{y ^{2}} - \frac{y ^{2}}{y ^{2}}) d y = \int (y^{- 2} - 1) d y$
Integrate term by term: $\frac{x}{y} = \frac{y ^{- 1}}{- 1} - y + C$
Simplify: $\frac{x}{y} = - \frac{1}{y} - y + C$

Finding the Constant $C$

Use initial condition: $x (1) = 1$ (This means when $y = 1$ , $x = 1$ )
Substitute $x = 1, y = 1$ into $\frac{x}{y} = - \frac{1}{y} - y + C$
$\frac{1}{1} = - \frac{1}{1} - 1 + C$
$1 = - 2 + C ⟹ C = 3$

The Particular Solution

Substitute $C = 3$ back: $\frac{x}{y} = - \frac{1}{y} - y + 3$
Multiply the entire equation by $y$ to isolate $x$ :
$x = - 1 - y^{2} + 3 y$

Evaluating $x (2)$

We need to find $x$ when $y = 2$ .
Substitute $y = 2$ into $x = - 1 - y^{2} + 3 y$
$x (2) = - 1 - (2)^{2} + 3 (2)$
$x (2) = - 1 - 4 + 6 = 1$

Final Calculation of $5 x (2)$

Target expression: $5 x (2)$
Substitute $x (2) = 1$ :
$5 (1) = 5$
Final Answer: 5

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The given differential equation is:

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

At first glance, it appears complex. However, we recognize this as a Linear Differential Equation where

x

is a function of

y

To reveal its standard form, we split the fraction on the right-hand side:

\frac{d x}{d y} = \frac{1}{y} + \frac{x}{y} - y

Next, we perform a strategic rearrangement by moving the

x

term to the left:

\frac{d x}{d y} - \frac{1}{y} x = \frac{1 - y ^{2}}{y}

This matches the standard linear form

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

, where

P (y) = - \frac{1}{y}

and

Q (y) = \frac{1 - y ^{2}}{y}

The Magic of the Integrating Factor

We now calculate the Integrating Factor (IF), which allows us to simplify the differential equation:

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

Since the integral of

- \frac{1}{y}

- ln (y)

, we have:

I F = e^{- l n (y)} = e^{l n (y^{- 1})} = \frac{1}{y}

This reduction to

\frac{1}{y}

significantly simplifies the subsequent integration process.

The Path to the Solution

We invoke the general solution formula

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

. Substituting our known values:

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

This simplifies to:

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

Integrating term by term, we obtain:

\frac{x}{y} = - \frac{1}{y} - y + C

To find the constant

C

, we apply the initial condition

x (1) = 1

. Substituting

y = 1

and

x = 1

1 = - 1 - 1 + C \Rightarrow C = 3

The Final Victory

Our particular solution is:

\frac{x}{y} = - \frac{1}{y} - y + 3

Multiplying by

y

to isolate

x

, we get:

x = - 1 - y^{2} + 3 y

To find

5 x (2)

, we first evaluate

x

y = 2

x (2) = - 1 - (2)^{2} + 3 (2) = - 1 - 4 + 6 = 1

Finally, multiplying by

5

, we arrive at the result:

$5 \cdot 1 = 5$

Common Traps & FAQs

Wait, why are we treating

x

as the dependent variable? Shouldn't it always be

y

That is a fantastic question! In calculus, we often get used to

y

being the dependent variable, but differential equations are flexible. The structure of the equation

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

dictates the relationship. By rearranging it, we see it fits the linear form perfectly with

x

as the function of

y

. Always let the equation guide your choice of dependent variable!

Why did the integrating factor

I F

become

1/ y

Great observation. We calculated

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

. The integral of

- 1/ y

- ln (y)

. So,

I F = e^{- l n (y)}

. Using the logarithmic property

a ln (b) = ln (b^{a})

, this becomes

e^{l n (y^{- 1})}

. Since

e^{l n (u)} = u

, we are left with

y^{- 1}

, which is

1/ y

. It is a beautiful simplification!

Why do we multiply the final equation by

y

at the end?

We do that to isolate

x

. Our solution at that stage was in the form

\frac{x}{y} = \dots

. To find the explicit function

x (y)

, we simply multiply both sides by

y

. It is just the final algebraic cleanup to make the function easy to evaluate for

x (2)

Animated Solution for Mathematics - Differential Equations: If $\frac{d x}{d y} = \frac{1 + x - y ^{2}}{y}, x (1) = 1$ , then $5 x (2)$ is equal to :

Enter Numerical Value:

Visualized Solution

Problem Statement

Converting to Standard Form

Identifying $P (y)$ and $Q (y)$

Integrating Factor (IF)

General Solution Setup

Evaluating the Integral

Finding the Constant $C$

The Particular Solution

Evaluating $x (2)$

Final Calculation of $5 x (2)$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Path to the Solution

The Final Victory

Similar Questions

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 :

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

If $x = x (t)$ is the solution of the differential equation $(t + 1) d x = (2 x + (t + 1)^{4}) d t, x (0) = 2$ , then, $x (1)$ equals

If a curve $y = f (x)$ passes through the point (1,2) and satisfies $x \frac{d y}{d x} + y = b x^{4}$ , then for what value of b, $\int_{1}^{2} f (x) d x = \frac{62}{5}$ ?

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

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 $y (3) = ?$

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 $y (x)$ 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:

Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dydx​=y1+x−y2​,x(1)=1, then 5x(2) is equal to :

Enter Numerical Value:

Visualized Solution

Problem Statement

Converting to Standard Form

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

Integrating Factor (IF)

General Solution Setup

Evaluating the Integral

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

The Particular Solution

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x(2)

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5x(2)

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Path to the Solution

The Final Victory

Similar Questions

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[1,∞)→[2,∞) be a differentiable function such that f(1)=2. If 6∫1x​f(t)dt=3xf(x)−x3 for all x≥1, then the value of f(2) is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=x(t) is the solution of the differential equation (t+1)dx=(2x+(t+1)4)dt,x(0)=2, then, x(1) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a curve y=f(x) passes through the point (1,2) and satisfies xdxdy​+y=bx4, then for what value of b, ∫12​f(x)dx=562​?

.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; } 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; } 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(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:

Animated Solution for Mathematics - Differential Equations: If $\frac{d x}{d y} = \frac{1 + x - y ^{2}}{y}, x (1) = 1$ , then $5 x (2)$ is equal to :

Identifying $P (y)$ and $Q (y)$

Finding the Constant $C$

Evaluating $x (2)$

Final Calculation of $5 x (2)$

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 :

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

If $x = x (t)$ is the solution of the differential equation $(t + 1) d x = (2 x + (t + 1)^{4}) d t, x (0) = 2$ , then, $x (1)$ equals

If a curve $y = f (x)$ passes through the point (1,2) and satisfies $x \frac{d y}{d x} + y = b x^{4}$ , then for what value of b, $\int_{1}^{2} f (x) d x = \frac{62}{5}$ ?

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

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 $y (3) = ?$

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 $y (x)$ 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: