Wait, why can't we just integrate $\frac{x+1}{x^2}$ as $\ln(x^2)$? That seems faster.

That is a very common trap! The rule $\int \frac{1}{u} du = \ln|u|$ only applies when the numerator is the derivative of the denominator. Here, the derivative of $x^2$ is $2x$, not $x+1$. By splitting the fraction into $\frac{1}{x} + \frac{1}{x^2}$, we treat each term with the correct integration rule: $\int \frac{1}{x} dx = \ln|x|$ and $\int x^{-2} dx = -x^{-1}$. Always simplify before you integrate!

JEE Main 2023 (29 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e$ . Then $lim_{x \to 0^{+}} f (x)$ is equal to

Q: Why is the limit $0$? Isn't $e^{-\infty}$ undefined?

Great question! In calculus, $e^{-\infty}$ is not undefined; it is a shorthand for $\lim_{u \to \infty} e^{-u}$, which is $\lim_{u \to \infty} \frac{1}{e^u}$. As $u$ grows larger and larger, the denominator $e^u$ explodes toward infinity, making the entire fraction shrink toward zero. So, $e^{-\infty}$ is effectively $0$.

Q: Why did we use $y(1)=e$ to find $C$ instead of just leaving it as a general solution?

The problem asks for the limit of $f(x)$, which is a specific function. A general solution with an arbitrary constant $C$ would lead to a family of curves. The initial condition $y(1)=e$ acts as a 'pin' that locks our curve to a specific path, allowing us to determine the exact value of $C$ and thus the unique function $f(x)$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Notice that the differential equation

y (x + 1) d x - x^{2} d y = 0

can be rearranged to isolate

d y

and

d x

terms on opposite sides. This is a classic separable differential equation. Visualize the variables

x

and

y

as two dancers that need to be separated before the music of integration can begin.

Hint 2

Hint 2: The Arsenal. Once you have

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

, do not panic at the fraction on the right. Use the algebraic property

\frac{x + 1}{x ^{2}} = \frac{x}{x ^{2}} + \frac{1}{x ^{2}}

to split it into two distinct, integrable terms:

\frac{1}{x}

and

x^{- 2}

Hint 3

Hint 3: The Execution. After integrating, you will find

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

. Use the initial condition

y (1) = e

to find the value of

C

. Once you have the specific function

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

, focus on the limit as

x \to 0^{+}

. Remember that as

x

approaches zero from the right,

\frac{1}{x}

approaches positive infinity, making the exponent approach negative infinity.

Visualized Solution

$A na l y z e t h eD i f f er e n t ia l E q u a t i o n$

Given differential equation: $y (x + 1) d x - x^{2} d y = 0$
Initial condition: $y (1) = e$
Goal: Find $f (x)$ and evaluate $lim_{x \to 0^{+}} f (x)$

$V a r iab l e S e p a r a t i o n$

Rearranging the equation: $x^{2} d y = y (x + 1) d x$
Separating variables: $\frac{d y}{y} = \frac{x + 1}{x ^{2}} d x$

$S im pl i f y t h e R H S$

Splitting the fraction: $\frac{d y}{y} = (\frac{x}{x ^{2}} + \frac{1}{x ^{2}}) d x$
Simplified form: $\frac{d y}{y} = (\frac{1}{x} + \frac{1}{x ^{2}}) d x$

$I n t e g r a t e B o t h S i d es$

Integrating: $\int \frac{d y}{y} = \int (\frac{1}{x} + \frac{1}{x ^{2}}) d x$
Result: $ln y = ln x - \frac{1}{x} + C$

$A ppl y I ni t ia l C o n d i t i o n$

Given $y (1) = e$ , substitute $x = 1, y = e$ :
$ln e = ln 1 - \frac{1}{1} + C$

$F in d t h e C o n s t an tC$

Evaluate logs: $1 = 0 - 1 + C$
Solve for $C$ : $C = 2$

$F or m t h e G e n er a l S o l u t i o n$

Substitute $C = 2$ : $ln y = ln x - \frac{1}{x} + 2$

$S o l v e f or y$

Taking exponential: $y = e^{l n x - \frac{1}{x} + 2}$
Simplifying: $y = e^{l n x} \cdot e^{2 - \frac{1}{x}}$
Final function: $y = x e^{2 - \frac{1}{x}}$

$S e t U pt h e L imi t$

Evaluate $lim_{x \to 0^{+}} x e^{2 - \frac{1}{x}}$

$A na l y z e t h e E x p o n e n t$

As $x \to 0^{+}$ , $\frac{1}{x} \to \infty$
Therefore, $2 - \frac{1}{x} \to - \infty$

$E v a l u a t e t h e E x p o n e n t ia l P a r t$

As exponent $\to - \infty$ , $e^{- \infty} \to 0$

$F ina l L imi tC a l c u l a t i o n$

Limit value: $0 \cdot 0 = 0$
Final Answer: $0$

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

We begin with the differential equation:

y (x + 1) d x - x^{2} d y = 0

This is a separable differential equation. Our first objective is to isolate the variables

x

and

y

on opposite sides of the equation.

Rearranging the terms, we obtain:

x^{2} d y = y (x + 1) d x

By dividing both sides by

x^{2} y

, we achieve the separation:

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

The Art of Simplification

Before integrating, we simplify the right-hand side by splitting the fraction into two distinct terms:

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

Substituting this back into our equation, we get:

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

Now, we perform the integration on both sides:

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

This yields the general solution:

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

Pinning the Curve

We are given the initial condition that the curve passes through the point

(1, e)

. Substituting

x = 1

and

y = e

into our general solution:

ln e = ln 1 - \frac{1}{1} + C

Since

ln e = 1

and

ln 1 = 0

, the equation simplifies to:

1 = 0 - 1 + C \Rightarrow C = 2

Thus, the specific solution is:

ln y = ln x - \frac{1}{x} + 2

To isolate

y

, we exponentiate both sides:

y = e^{l n x - \frac{1}{x} + 2}

Using the properties of exponents, we simplify this to:

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

The Climax

Approaching the Origin

Finally, we evaluate the limit as

x

approaches

0

from the positive side:

x \to 0^{+} lim x e^{2 - \frac{1}{x}}

x \to 0^{+}

, the term

\frac{1}{x} \to \infty

. Consequently, the exponent

2 - \frac{1}{x} \to - \infty

Since

e^{- \infty} \to 0

, the expression becomes:

x \to 0^{+} lim (x \cdot 0) = 0

The final result of the limit is 0.

Common Traps & FAQs

Wait, why can't we just integrate

\frac{x + 1}{x ^{2}}

ln (x^{2})

? That seems faster.

That is a very common trap! The rule

\int \frac{1}{u} d u = ln ∣ u ∣

only applies when the numerator is the derivative of the denominator. Here, the derivative of

x^{2}

2 x

, not

x + 1

. By splitting the fraction into

\frac{1}{x} + \frac{1}{x ^{2}}

, we treat each term with the correct integration rule:

\int \frac{1}{x} d x = ln ∣ x ∣

and

\int x^{- 2} d x = - x^{- 1}

. Always simplify before you integrate!

Why is the limit

0

? Isn't

e^{- \infty}

undefined?

Great question! In calculus,

e^{- \infty}

is not undefined; it is a shorthand for

lim_{u \to \infty} e^{- u}

, which is

lim_{u \to \infty} \frac{1}{e ^{u}}

. As

u

grows larger and larger, the denominator

e^{u}

explodes toward infinity, making the entire fraction shrink toward zero. So,

e^{- \infty}

is effectively

0

Why did we use

y (1) = e

to find

C

instead of just leaving it as a general solution?

The problem asks for the limit of

f (x)

, which is a specific function. A general solution with an arbitrary constant

C

would lead to a family of curves. The initial condition

y (1) = e

acts as a 'pin' that locks our curve to a specific path, allowing us to determine the exact value of

C

and thus the unique function

f (x)

Animated Solution for Mathematics - Differential Equations: Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e$ . Then $lim_{x \to 0^{+}} f (x)$ is equal to

Select Answer:

Visualized Solution

$A na l y z e t h eD i f f er e n t ia l E q u a t i o n$

$V a r iab l e S e p a r a t i o n$

$S im pl i f y t h e R H S$

$I n t e g r a t e B o t h S i d es$

$A ppl y I ni t ia l C o n d i t i o n$

$F in d t h e C o n s t an tC$

$F or m t h e G e n er a l S o l u t i o n$

$S o l v e f or y$

$S e t U pt h e L imi t$

$A na l y z e t h e E x p o n e n t$

$E v a l u a t e t h e E x p o n e n t ia l P a r t$

$F ina l L imi tC a l c u l a t i o n$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Simplification

Pinning the Curve

The Climax

Similar Questions

If $y = y (x)$ 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 $x = x (y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that $x (1) = 0$ . Then, $x (e)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0, y (0) = 1$ . Then $y (\frac{π}{4})$ is equal to

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

Let $f : R \to R$ be a differentiable function with $f (0) = 0$ . If $y = f (x)$ satisfies the differential equation $\frac{d y}{d x} = (2 + 5 y) (y - 2)$ , then the value of $lim_{x \to \infty} f (x)$ is ________.

The solution of the differential equation $\frac{d y}{d x} = (x - y)^{2}$ , when $y (1) = 1$ , is:

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

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

Which of the following is true for $y (x)$ that satisfies the differential equation $\frac{d y}{d x} = x y - 1 + x - y; y (0) = 0$

Select Answer:

Visualized Solution

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

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

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

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

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

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

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

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

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

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

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

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

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Simplification

Pinning the Curve

The Climax

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 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 x=x(y) be the solution of the differential equation 2yex/y2dx+(y2−4xex/y2)dy=0 such that x(1)=0. Then, x(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 (1+y2)etanxdx+cos2x(1+e2tanx)dy=0,y(0)=1. Then y(4π​) 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 differential equation (1+e2x)dxdy​+2(1+y2)ex=0 and y(0)=0, then 6[y′(0)+(y(loge​3​))2] is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a differentiable function with f(0)=0. If y=f(x) satisfies the differential equation dxdy​=(2+5y)(y−2), then the value of limx→∞​f(x) is ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the differential equation dxdy​=(x−y)2, when y(1)=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 ex1−y2​dx+(xy​)dy=0,y(1)=−1. Then the value of (y(3))2 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the solution curve y=y(x) of the differential equation (1+y2)(1+loge​x)dx+xdy=0,x>0 passes through the point (1,1) and y(e)=β+tan(3/2)α−tan(3/2)​, then α+2β is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ∫0x​1−(y′(t))2​dt=∫0x​y(t)dt,0≤x≤3,y≥0,y(0)=0. Then at x=2,y′′+y+1 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following is true for y(x) that satisfies the differential equation dxdy​=xy−1+x−y;y(0)=0

$A na l y z e t h eD i f f er e n t ia l E q u a t i o n$

$V a r iab l e S e p a r a t i o n$

$S im pl i f y t h e R H S$

$I n t e g r a t e B o t h S i d es$

$A ppl y I ni t ia l C o n d i t i o n$

$F in d t h e C o n s t an tC$

$F or m t h e G e n er a l S o l u t i o n$

$S o l v e f or y$

$S e t U pt h e L imi t$

$A na l y z e t h e E x p o n e n t$

$E v a l u a t e t h e E x p o n e n t ia l P a r t$

$F ina l L imi tC a l c u l a t i o n$

If $y = y (x)$ 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 $x = x (y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that $x (1) = 0$ . Then, $x (e)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0, y (0) = 1$ . Then $y (\frac{π}{4})$ is equal to

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

Let $f : R \to R$ be a differentiable function with $f (0) = 0$ . If $y = f (x)$ satisfies the differential equation $\frac{d y}{d x} = (2 + 5 y) (y - 2)$ , then the value of $lim_{x \to \infty} f (x)$ is ________.

The solution of the differential equation $\frac{d y}{d x} = (x - y)^{2}$ , when $y (1) = 1$ , is:

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

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

Which of the following is true for $y (x)$ that satisfies the differential equation $\frac{d y}{d x} = x y - 1 + x - y; y (0) = 0$