JEE Main 2021 (February)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is $\frac{x ^{2} - 4 x + y + 8}{x - 2}$ , then this curve also passes through the point :

Q: What if the constant $C$ wasn't zero? Does that change the curve?

Absolutely! The constant $C$ represents the family of curves that satisfy this differential equation. By using the initial condition that the curve passes through the origin $(0,0)$, we specifically selected the one unique curve from that family that passes through that point. Without that condition, we would have an infinite set of curves!

Select Answer:

Visualized Solution

Defining the Slope of the Tangent

Slope of the tangent at any point $(x, y)$ is given by $\frac{d y}{d x}$
$\frac{d y}{d x} = \frac{x ^{2} - 4 x + y + 8}{x - 2}$

Algebraic Simplification

Notice the denominator is $(x - 2)$
Rewrite the numerator to create $(x - 2)^{2}$ terms: $x^{2} - 4 x + 8 + y = (x^{2} - 4 x + 4) + 4 + y$
$\frac{d y}{d x} = \frac{( x - 2 ) ^{2} + ( y + 4 )}{x - 2}$

Splitting the Fraction

Separate the terms in the numerator over the denominator
$\frac{d y}{d x} = \frac{( x - 2 ) ^{2}}{x - 2} + \frac{y + 4}{x - 2}$
$\frac{d y}{d x} = (x - 2) + \frac{y + 4}{x - 2}$

Substitution for Simplification

Let $x - 2 = t ⟹ d x = d t$
Let $y + 4 = u ⟹ d y = d u$
Substitute these into the differential equation:
$\frac{d u}{d t} = t + \frac{u}{t}$

Forming the Linear Differential Equation

Rearrange the terms: $\frac{d u}{d t} - \frac{1}{t} u = t$
This matches the standard Linear Differential Equation (LDE) form: $\frac{d u}{d t} + P (t) u = Q (t)$
Here, $P (t) = - \frac{1}{t}$ and $Q (t) = t$

Calculating the Integrating Factor

Integrating Factor (I.F.) $= e^{\int P (t) d t}$
I.F. $= e^{\int - \frac{1}{t} d t} = e^{- l n ∣ t ∣}$
Using logarithm properties: $e^{l n ∣ t^{- 1} ∣} = \frac{1}{t}$

Solving the Differential Equation

The general solution is: $u \cdot (I.F.) = \int Q (t) \cdot (I.F.) d t$
Substitute the knowns: $u \cdot \frac{1}{t} = \int t \cdot \frac{1}{t} d t$
Simplify the integral: $\frac{u}{t} = \int 1 d t$
$\frac{u}{t} = t + C$

Back-Substitution

Replace $u$ and $t$ with original variables $x$ and $y$
Recall: $u = y + 4$ and $t = x - 2$
$\frac{y + 4}{x - 2} = (x - 2) + C$

Applying the Initial Condition

The problem states the curve passes through the origin $(0, 0)$
Substitute $x = 0, y = 0$ into the equation:
$\frac{0 + 4}{0 - 2} = (0 - 2) + C$
$- 2 = - 2 + C ⟹ C = 0$

The Final Equation of the Curve

Substitute $C = 0$ back into the equation
$\frac{y + 4}{x - 2} = x - 2$
Multiply both sides by $(x - 2)$ :
$y + 4 = (x - 2)^{2}$

Verifying the Given Options

We need to find which point lies on $y + 4 = (x - 2)^{2}$
Let's test the option $(5, 5)$ :
LHS: $y + 4 = 5 + 4 = 9$
RHS: $(x - 2)^{2} = (5 - 2)^{2} = 3^{2} = 9$
Since LHS = RHS, the curve passes through $(5, 5)$

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The problem provides the differential equation for the slope of a curve at any point

(x, y)

\frac{d y}{d x} = \frac{x ^{2} - 4 x + y + 8}{x - 2}

To simplify the numerator, we recognize that

x^{2} - 4 x

can be completed into a perfect square. By adding and subtracting

4

, we rewrite the expression as:

x^{2} - 4 x + y + 8 = (x^{2} - 4 x + 4) + y + 4 = (x - 2)^{2} + (y + 4)

Simplifying the Equation

Substituting this back into our original differential equation, we obtain:

\frac{d y}{d x} = \frac{( x - 2 ) ^{2} + ( y + 4 )}{x - 2}

By splitting the fraction, the equation simplifies to:

\frac{d y}{d x} = (x - 2) + \frac{y + 4}{x - 2}

Change of Variables

To solve this, we introduce the substitutions

t = x - 2

and

u = y + 4

. Consequently,

d t = d x

and

d u = d y

. The equation transforms into:

\frac{d u}{d t} = t + \frac{u}{t}

Rearranging this into the standard form of a Linear Differential Equation, we get:

\frac{d u}{d t} - \frac{1}{t} u = t

Solving the Linear Differential Equation

The integrating factor (

I . F .

) is calculated as follows:

I . F . = e^{\int P (t) d t} = e^{\int - \frac{1}{t} d t} = e^{- l n ∣ t ∣} = \frac{1}{t}

Multiplying the entire equation by the

I . F .

, we have:

\frac{1}{t} \frac{d u}{d t} - \frac{1}{t ^{2}} u = 1

The left side is the derivative of the product

\frac{u}{t}

\frac{d}{d t} (\frac{u}{t}) = 1

Final Calculation

Integrating both sides with respect to

t

yields:

\frac{u}{t} = t + C

Substituting back

u = y + 4

and

t = x - 2

, we arrive at the general solution:

\frac{y + 4}{x - 2} = x - 2 + C

Given that the curve passes through the origin

(0, 0)

, we substitute

x = 0

and

y = 0

to find

C

\frac{0 + 4}{0 - 2} = 0 - 2 + C \Rightarrow - 2 = - 2 + C \Rightarrow C = 0

The final equation of the curve is:

$y + 4 = (x - 2)^{2}$

Common Traps & FAQs

Wait, why can't we just integrate the slope directly to get the curve?

That is a great question! We can only integrate directly if the variables are separable, meaning we can get all the

x

terms on one side and all the

y

terms on the other. In this case, the

y

term is trapped in the numerator with

x

terms, so direct integration is impossible. We must use the tools of differential equations to isolate the relationship.

How did you know to substitute

x - 2

and

y + 4

Pattern recognition is a superpower in calculus! When you see a denominator like

(x - 2)

and a numerator that looks like a quadratic, it is a huge hint to try and express the numerator in terms of that denominator. By completing the square for

x^{2} - 4 x + 4

, we created the

(x - 2)^{2}

term, which made the substitution obvious and necessary.

What if the constant

C

wasn't zero? Does that change the curve?

Absolutely! The constant

C

represents the family of curves that satisfy this differential equation. By using the initial condition that the curve passes through the origin

(0, 0)

, we specifically selected the one unique curve from that family that passes through that point. Without that condition, we would have an infinite set of curves!

Animated Solution for Mathematics - Differential Equations: If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is $\frac{x ^{2} - 4 x + y + 8}{x - 2}$ , then this curve also passes through the point :

Select Answer:

Visualized Solution

Defining the Slope of the Tangent

Algebraic Simplification

Splitting the Fraction

Substitution for Simplification

Forming the Linear Differential Equation

Calculating the Integrating Factor

Solving the Differential Equation

Back-Substitution

Applying the Initial Condition

The Final Equation of the Curve

Verifying the Given Options

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Simplifying the Equation

Change of Variables

Solving the Linear Differential Equation

Final Calculation

Similar Questions

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Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If $y (0) = 0$ , then $y (2)$ is equal to

Select Answer:

Visualized Solution

Defining the Slope of the Tangent

Algebraic Simplification

Splitting the Fraction

Substitution for Simplification

Forming the Linear Differential Equation

Calculating the Integrating Factor

Solving the Differential Equation

Back-Substitution

Applying the Initial Condition

The Final Equation of the Curve

Verifying the Given Options

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Simplifying the Equation

Change of Variables

Solving the Linear Differential Equation

Final Calculation

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