JEE Main 2018 (15 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: 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 :

Q: Is there a shortcut? Can I just integrate $f(x)$ directly?

I understand the temptation! However, $f(x)$ is not the solution; it is the 'source' or 'driving force' in the differential equation. You cannot simply integrate $f(x)$ because $y$ is coupled with its derivative $\frac{dy}{dx}$. You must follow the standard procedure for linear differential equations to account for how the system responds to that driving force.

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the function

f (x)

as a physical switch. It is not a single continuous function but two distinct regimes. The differential equation changes its nature at

x = 1

. You must solve it in two separate phases: one for the interval

[0, 1]

and another for

x > 1

Hint 2

The Arsenal: This is a First-Order Linear Differential Equation of the form

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

. Your primary tool is the Integrating Factor, defined as

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

. Since

P = 2

, your I.F. is

e^{2 x}

. Multiply the entire equation by this factor to make the left side a derivative of a product.

Hint 3

The Execution: For

x \in [0, 1]

, solve

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

using the initial condition

y (0) = 0

to find the first constant

C_{1}

. For

x > 1

, solve

\frac{d y}{d x} + 2 y = 0

to get

y = C_{2} e^{- 2 x}

. Finally, use the continuity condition

y (1)_{l e f t} = y (1)_{r i g h t}

to determine

C_{2}

and calculate

y (3/2)

Visualized Solution

Identifying the Differential Equation

Given DE: $\frac{d y}{d x} + 2 y = f (x)$
This is a First-Order Linear Differential Equation of the form $\frac{d y}{d x} + P y = Q$ .
Here, $P = 2$ and $Q = f (x)$ .
$f (x)$ is piecewise: $1$ for $x \in [0, 1]$ and $0$ otherwise.

Calculating the Integrating Factor

To solve this, we need the Integrating Factor (I.F.).
Formula: $I.F. = e^{\int P d x}$
Substitute $P = 2$ :
$I.F. = e^{\int 2 d x} = e^{2 x}$

Solving for $x \in [0, 1]$

For the interval $x \in [0, 1]$ , the function $f (x) = 1$ .
The general solution formula is: $y \cdot (I.F.) = \int Q \cdot (I.F.) d x + C_{1}$
Substituting the values:
$y e^{2 x} = \int 1 \cdot e^{2 x} d x + C_{1}$
$y e^{2 x} = \frac{e ^{2 x}}{2} + C_{1}$

Applying Initial Condition $y (0) = 0$

We are given the initial condition $y (0) = 0$ .
Substitute $x = 0$ and $y = 0$ into our equation:
$0 \cdot e^{0} = \frac{e ^{0}}{2} + C_{1}$
$0 = \frac{1}{2} + C_{1} ⟹ C_{1} = - \frac{1}{2}$
The solution for $x \in [0, 1]$ becomes: $y e^{2 x} = \frac{e ^{2 x} - 1}{2}$
$y (x) = \frac{1 - e ^{- 2 x}}{2}$

Finding the Boundary Value at $x = 1$

The solution curve $y (x)$ must be continuous at the boundary $x = 1$ .
Let's find the value of $y$ at this exact point:
$y (1) = \frac{1 - e ^{- 2 (1)}}{2}$
$y (1) = \frac{1 - e ^{- 2}}{2} = \frac{e ^{2} - 1}{2 e ^{2}}$

Solving for $x > 1$

Now consider the interval $x > 1$ , where $f (x) = 0$ .
The differential equation simplifies to: $\frac{d y}{d x} + 2 y = 0$
Using the same Integrating Factor $e^{2 x}$ :
$y e^{2 x} = \int 0 \cdot e^{2 x} d x + C_{2}$
$y e^{2 x} = C_{2} ⟹ y (x) = C_{2} e^{- 2 x}$

Determining Constant $C_{2}$ via Continuity

For the function to be continuous at $x = 1$ , the left-hand limit must equal the right-hand limit.
From the second part, at $x = 1$ : $y (1) = C_{2} e^{- 2 (1)} = C_{2} e^{- 2}$
Equating this with our previously found boundary value:
$C_{2} e^{- 2} = \frac{e ^{2} - 1}{2 e ^{2}}$
Multiply both sides by $e^{2}$ :
$C_{2} = \frac{e ^{2} - 1}{2}$

Final Calculation for $y (3/2)$

We need to find the value of $y (3/2)$ .
Since $3/2 > 1$ , we use the solution for the second interval: $y (x) = C_{2} e^{- 2 x}$
Substitute $x = 3/2$ and $C_{2} = \frac{e ^{2} - 1}{2}$ :
$y (3/2) = (\frac{e ^{2} - 1}{2}) e^{- 2 (3/2)}$
$y (3/2) = \frac{e ^{2} - 1}{2} e^{- 3}$
$y (3/2) = \frac{e ^{2} - 1}{2 e ^{3}}$

Conclusion and Key Takeaway

Final Answer: $y (3/2) = \frac{e ^{2} - 1}{2 e ^{3}}$
Key Takeaway: For differential equations with piecewise functions, solve them in segments.
Crucial Step: Always use continuity at the boundary points to link the constants of integration.
This ensures the physical or geometric solution has no breaks.

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The motion of the particle is governed by the differential equation:

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

The field

f (x)

is defined piecewise:

f (x) = {1, 0, 0 \leq x \leq 1 x > 1

To solve this, we must treat the motion as a two-act process, ensuring continuity at the boundary

x = 1

Act I

The Driven Motion

For the interval

x \in [0, 1]

, the equation is

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

. We utilize the Integrating Factor method, where the I.F. is:

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

Multiplying the differential equation by

e^{2 x}

yields:

\frac{d}{d x} (y e^{2 x}) = e^{2 x}

Integrating both sides, we obtain

y e^{2 x} = \frac{e ^{2 x}}{2} + C_{1}

. Applying the initial condition

y (0) = 0

0 = \frac{1}{2} + C_{1} \Rightarrow C_{1} = - \frac{1}{2}

Thus, for

x \in [0, 1]

, the solution is:

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

The Bridge of Continuity

To ensure the particle's path is continuous, we calculate the position at the boundary

x = 1

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

This value serves as the initial condition for the second phase of the motion.

Act II

The Free Decay

For

x > 1

, the field vanishes, resulting in the homogeneous equation

\frac{d y}{d x} + 2 y = 0

. The general solution is:

y (x) = C_{2} e^{- 2 x}

We enforce continuity at

x = 1

by setting the two solutions equal:

C_{2} e^{- 2 (1)} = \frac{e ^{2} - 1}{2 e ^{2}}

Multiplying both sides by

e^{2}

, we isolate the constant:

C_{2} = \frac{e ^{2} - 1}{2}

The Final Reveal

We are tasked with finding the value of

y

x = 3/2

. Since

3/2 > 1

, we use the second-act solution:

y (3/2) = C_{2} e^{- 2 (3/2)} = C_{2} e^{- 3}

Substituting the value of

C_{2}

derived previously:

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

The final result is:

$y (3/2) = \frac{e ^{2} - 1}{2 e ^{3}}$

Common Traps & FAQs

Wait, why do we need to use continuity at

x = 1

? Can't we just solve the two equations independently?

Great question! While the differential equations are independent, the solution

y (x)

represents a single, continuous physical process. If the function were discontinuous at

x = 1

, the derivative

\frac{d y}{d x}

would be undefined (or infinite) at that point, which contradicts the existence of a solution to the differential equation. Continuity is the 'glue' that connects the two segments.

Why does the constant

C_{1}

disappear when we move to the second interval?

It doesn't disappear; it is simply part of the first segment's solution. When we move to the interval

x > 1

, the differential equation changes because the driving force

f (x)

changes. This creates a new general solution with its own constant,

C_{2}

. We use the boundary value from the first segment to fix the value of

C_{2}

Is there a shortcut? Can I just integrate

f (x)

directly?

I understand the temptation! However,

f (x)

is not the solution; it is the 'source' or 'driving force' in the differential equation. You cannot simply integrate

f (x)

because

y

is coupled with its derivative

\frac{d y}{d x}

. You must follow the standard procedure for linear differential equations to account for how the system responds to that driving force.

Animated Solution for Mathematics - Differential Equations: 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 :

Select Answer:

Visualized Solution

Identifying the Differential Equation

Calculating the Integrating Factor

Solving for $x \in [0, 1]$

Applying Initial Condition $y (0) = 0$

Finding the Boundary Value at $x = 1$

Solving for $x > 1$

Determining Constant $C_{2}$ via Continuity

Final Calculation for $y (3/2)$

Conclusion and Key Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Act I

The Bridge of Continuity

Act II

The Final Reveal

Similar Questions

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:

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 $y = y (x)$ 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 $y = y (x)$ 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 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 :

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 $y = y (x)$ 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 $y = y (x)$ be the solution of the differential equation $(2 x lo g_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} lo g_{e} x, x > 0$ and $y (e^{- 1}) = 0$ . Then, $y (e)$ is equal to

If $x = f (y)$ is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{t a n^{- 1} y}) \frac{d y}{d x} = 0,$ $y \in (- \frac{π}{2}, \frac{π}{2})$ with $f (0) = 1$ , then $f (\frac{1}{3})$ is equal to:

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 $y (3) = ?$

Select Answer:

Visualized Solution

Identifying the Differential Equation

Calculating the Integrating Factor

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[0,1]

Applying Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(0)=0

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

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

Determining Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​ via Continuity

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(3/2)

Conclusion and Key Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Act I

The Bridge of Continuity

Act II

The Final Reveal

Similar Questions

.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; } 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=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=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 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; } 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 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 y=y(x) be the solution of the differential equation (2xloge​x)dxdy​+2y=x3​loge​x,x>0 and y(e−1)=0. Then, y(e) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=f(y) is the solution of the differential equation (1+y2)+(x−2etan−1y)dxdy​=0, y∈(−2π​,2π​) with f(0)=1, then f(3​1​) is equal to:

.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)=?

Solving for $x \in [0, 1]$

Applying Initial Condition $y (0) = 0$

Finding the Boundary Value at $x = 1$

Solving for $x > 1$

Determining Constant $C_{2}$ via Continuity

Final Calculation for $y (3/2)$

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:

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 $y = y (x)$ 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 $y = y (x)$ 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 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 :

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 $y = y (x)$ 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 $y = y (x)$ be the solution of the differential equation $(2 x lo g_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} lo g_{e} x, x > 0$ and $y (e^{- 1}) = 0$ . Then, $y (e)$ is equal to

If $x = f (y)$ is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{t a n^{- 1} y}) \frac{d y}{d x} = 0,$ $y \in (- \frac{π}{2}, \frac{π}{2})$ with $f (0) = 1$ , then $f (\frac{1}{3})$ is equal to:

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 $y (3) = ?$