JEE Main 2022 (27 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y_{1} (x)$ and $y = y_{2} (x)$ be two distinct solutions of the differential equation $\frac{d y}{d x} = x + y$ , with $y_{1} (0) = 0$ and $y_{2} (0) = 1$ respectively. Then, the number of points of intersection of $y = y_{1} (x)$ and $y = y_{2} (x)$ is

Select Answer:

Progressive Hints

Hint 1

Visualize the differential equation

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

not just as an algebraic expression, but as a slope field. Every point

(x, y)

in the Cartesian plane dictates the direction of the solution curve. Ask yourself: if two curves start at different heights at

x = 0

, can they ever cross without violating the uniqueness of the solution at the intersection point?

Hint 2

Recall the standard linear differential equation form:

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

. Your toolkit must include the Integrating Factor,

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

. This is the bridge that transforms the left side of your equation into the derivative of a product, making the integration possible.

Hint 3

Once you derive the general solution

y = C e^{x} - x - 1

, set the two specific solutions equal to each other:

y_{1} (x) = y_{2} (x)

. Watch what happens to the terms involving

x

and the constant

1

. The cancellation will reveal the truth about their intersection.

Visualized Solution

Standard Linear Form

Given equation: $\frac{d y}{d x} = x + y$
Rearranging to standard linear form: $\frac{d y}{d x} - y = x$
Standard Form: $\frac{d y}{d x} + P (x) y = Q (x)$
Comparing terms: $P (x) = - 1$ and $Q (x) = x$

Integrating Factor

Integrating Factor (I.F.) formula: $e^{\int P (x) d x}$
Substitute $P (x) = - 1$ : $I . F . = e^{\int - 1 d x}$
Evaluating the integral: $I . F . = e^{- x}$

General Solution Setup

General solution formula: $y (I . F .) = \int Q (x) (I . F .) d x + C$
Substituting values: $y e^{- x} = \int x e^{- x} d x + C$

Integration by Parts

Using $\int uv d x = u \int v d x - \int (u^{'} \int v d x) d x$
Let $u = x$ and $v = e^{- x}$
$\int x e^{- x} d x = x (- e^{- x}) - \int (1) (- e^{- x}) d x$
$= - x e^{- x} + \int e^{- x} d x$
$= - x e^{- x} - e^{- x}$

General Solution for $y$

Substitute integral back: $y e^{- x} = - x e^{- x} - e^{- x} + C$
Multiply by $e^{x}$ : $y = - x - 1 + C e^{x}$

First Solution $y_{1} (x)$

For $y_{1} (x)$ , given initial condition $y_{1} (0) = 0$
Substitute $x = 0, y = 0$ : $0 = - 0 - 1 + C_{1} e^{0}$
$0 = - 1 + C_{1} ⟹ C_{1} = 1$
Solution $y_{1} (x) = e^{x} - x - 1$

Second Solution $y_{2} (x)$

For $y_{2} (x)$ , given initial condition $y_{2} (0) = 1$
Substitute $x = 0, y = 1$ : $1 = - 0 - 1 + C_{2} e^{0}$
$1 = - 1 + C_{2} ⟹ C_{2} = 2$
Solution $y_{2} (x) = 2 e^{x} - x - 1$

Intersection Condition

To find intersection points, set $y_{1} (x) = y_{2} (x)$
$e^{x} - x - 1 = 2 e^{x} - x - 1$

Solving for Intersection

Simplifying the equation: $e^{x} = 2 e^{x}$
Subtracting $e^{x}$ from both sides: $e^{x} = 0$
Since $e^{x} > 0$ for all real $x$ , this equation has no solution.

Conclusion

The curves $y = y_{1} (x)$ and $y = y_{2} (x)$ do not intersect.
Number of points of intersection = 0
Key takeaway: Distinct solutions of a first-order linear DE with the same $P (x)$ and $Q (x)$ cannot intersect due to the existence and uniqueness theorem.

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

We are tasked with solving the differential equation

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

. This equation describes a relationship between the slope of a curve and its coordinates.

To solve this, we first transform it into the standard linear form:

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

This matches the classic form

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

, where

P (x) = - 1

and

Q (x) = x

The Magic of the Integrating Factor

To solve this linear equation, we utilize an Integrating Factor (

I . F .

), defined as

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

. Substituting

P (x) = - 1

, we calculate:

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

Multiplying the entire differential equation by

e^{- x}

allows us to express the left side as the derivative of a product:

\frac{d}{d x} (y e^{- x}) = x e^{- x}

Integrating both sides with respect to

x

, we obtain:

y e^{- x} = \int x e^{- x} d x + C

The Art of Integration by Parts

To evaluate the integral

\int x e^{- x} d x

, we apply the technique of integration by parts. Setting

u = x

and

d v = e^{- x} d x

, we use the formula

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

\int x e^{- x} d x = x (- e^{- x}) - \int (- e^{- x}) d x = - x e^{- x} - e^{- x}

Combining this result with our constant of integration

C

, we have:

y e^{- x} = - x e^{- x} - e^{- x} + C

Multiplying the entire equation by

e^{x}

yields the general solution:

y = - x - 1 + C e^{x}

The Moment of Truth

Intersection

We now determine the specific curves

y_{1} (x)

and

y_{2} (x)

based on the given initial conditions.

For

y_{1} (0) = 0

0 = - 0 - 1 + C_{1} (1) \Rightarrow C_{1} = 1 \Rightarrow y_{1} (x) = e^{x} - x - 1

For

y_{2} (0) = 1

1 = - 0 - 1 + C_{2} (1) \Rightarrow C_{2} = 2 \Rightarrow y_{2} (x) = 2 e^{x} - x - 1

To find the points of intersection, we set

y_{1} (x) = y_{2} (x)

e^{x} - x - 1 = 2 e^{x} - x - 1

Simplifying the equation, the terms

- x

and

- 1

cancel out, leaving:

e^{x} = 2 e^{x} \Rightarrow e^{x} = 0

Since the exponential function

e^{x}

is strictly positive for all real

x

, it can never equal zero. Therefore, the two curves never intersect. The number of points of intersection is 0.

Common Traps & FAQs

Wait, why can't we just use energy conservation or some other shortcut to find the intersection?

That is a brilliant instinct, but differential equations are fundamentally about local slopes, not global energy states. Energy conservation is a tool for systems where forces are conservative and work is path-independent. Here, we are dealing with the 'flow' of a function. The only shortcut here is the Uniqueness Theorem, which tells us that for a first-order linear differential equation, two distinct solutions cannot intersect. If they did, the solution at that intersection point would not be unique, which contradicts the theorem.

Can

e^{x}

ever be zero? I feel like I missed a step.

You haven't missed a step; you've hit a fundamental truth of calculus! The exponential function

e^{x}

is strictly positive for all real values of

x

. It approaches zero as

x

goes to negative infinity, but it never actually touches the x-axis. If your algebra leads you to

e^{x} = 0

, it is a mathematical signal that the condition you are testing—the intersection—is impossible.

Is there a formal way to prove this without solving the equation?

Yes, and this is the 'pro' level of JEE preparation. The Picard-Lindelöf theorem (or the Existence and Uniqueness Theorem for linear differential equations) guarantees that if

P (x)

and

Q (x)

are continuous on an interval, then for any initial condition

y (x_{0}) = y_{0}

, there exists a unique solution. Since our two curves have different initial conditions (

y_{1} (0) = 0

and

y_{2} (0) = 1

), they are distinct solutions. By the theorem, they cannot intersect. Solving the equation is just the 'verification' of this beautiful theoretical fact.

Select Answer:

Visualized Solution

Standard Linear Form

Integrating Factor

General Solution Setup

Integration by Parts

General Solution for $y$

First Solution $y_{1} (x)$

Second Solution $y_{2} (x)$

Intersection Condition

Solving for Intersection

Conclusion

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Art of Integration by Parts

The Moment of Truth

Similar Questions

A solution curve of the differential equation $(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0, x > 0$ , passes through the point (1, 3). Then the solution curve

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 :

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 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 $y (3) = ?$

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to

Let $y^{'} (x) + y (x) g^{'} (x) = g (x) g^{'} (x)$ , $y (0) = 0$ , $x \in R$ , where $f^{'} (x)$ denotes $\frac{df ( x )}{d x}$ and $g (x)$ is a given non-constant differentiable function on $R$ with $g (0) = g (2) = 0$ . Then the value of $y (2)$ is

A solution of the differential equation $(\frac{d y}{d x})^{2} - x \frac{d y}{d x} + y = 0$ is

Let $α x = exp (x^{β} y^{γ})$ be the solution of the differential equation $2 x^{2} y d y - (1 - x y^{2}) d x = 0, x > 0, y (2) = lo g_{e} 2$ . Then $α + β - γ$ equals :

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 :

If $y = y (x)$ is the solution of the equation $e^{s i n y} cos y \frac{d y}{d x} + e^{s i n y} cos x = cos x, y (0) = 0$ ; then $1 + y (\frac{π}{2}) + \frac{π}{2} y^{'} (\frac{π}{2}) + \frac{1}{2} y^{''} (\frac{π}{2})$ is equal to

Select Answer:

Visualized Solution

Standard Linear Form

Integrating Factor

General Solution Setup

Integration by Parts

General Solution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

First Solution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y1​(x)

Second Solution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2​(x)

Intersection Condition

Solving for Intersection

Conclusion

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Art of Integration by Parts

The Moment of Truth

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A solution curve of the differential equation (x2+xy+4x+2y+4)dxdy​−y2=0,x>0, passes through the point (1, 3). Then the solution curve

.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; } 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 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; } Let y=y(x) be the solution of the differential equation 2 cos xdxdy​=sin 2x−4y sin x,x∈(0,2π​). If y(3π​)=0 then y′(4π​)+y(4π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A solution of the differential equation (dxdy​)2−xdxdy​+y=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let αx=exp(xβyγ) be the solution of the differential equation 2x2ydy−(1−xy2)dx=0,x>0,y(2)=loge​2​. Then α+β−γ equals :

.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; } If y=y(x) is the solution of the equation esinycosydxdy​+esinycosx=cosx,y(0)=0; then 1+y(2π​)+2π​​y′(2π​)+2​1​y′′(2π​) is equal to

General Solution for $y$

First Solution $y_{1} (x)$

Second Solution $y_{2} (x)$

A solution curve of the differential equation $(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0, x > 0$ , passes through the point (1, 3). Then the solution curve

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 :

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 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 $y (3) = ?$

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to

A solution of the differential equation $(\frac{d y}{d x})^{2} - x \frac{d y}{d x} + y = 0$ is

Let $α x = exp (x^{β} y^{γ})$ be the solution of the differential equation $2 x^{2} y d y - (1 - x y^{2}) d x = 0, x > 0, y (2) = lo g_{e} 2$ . Then $α + β - γ$ equals :

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 :

If $y = y (x)$ is the solution of the equation $e^{s i n y} cos y \frac{d y}{d x} + e^{s i n y} cos x = cos x, y (0) = 0$ ; then $1 + y (\frac{π}{2}) + \frac{π}{2} y^{'} (\frac{π}{2}) + \frac{1}{2} y^{''} (\frac{π}{2})$ is equal to