JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: 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 :

Q: Why do we add the constant $C$ only on the right side?

In the general solution $y \cdot IF = \int Q(x) \cdot IF dx + C$, the constant $C$ is added after performing the integration on the right side. It represents the family of curves that satisfy the differential equation. We then use the initial condition $(1, 0)$ to find the specific value of $C$ for our unique curve.

Select Answer:

Visualized Solution

The Given Differential Equation

Given DE: $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0$
Divide by $d x$ : $x (x^{2} + e^{x}) \frac{d y}{d x} + e^{x} (x - 2) y - x^{3} = 0$

Rearranging to Standard Linear Form

Rearranging: $\frac{d y}{d x} + \frac{e ^{x} ( x - 2 )}{x ( x ^{2} + e ^{x} )} y = \frac{x ^{2}}{x ^{2} + e ^{x}}$
This is of the form: $\frac{d y}{d x} + P (x) y = Q (x)$

Identifying $P (x)$ and $Q (x)$

$P (x) = \frac{e ^{x} ( x - 2 )}{x ( x ^{2} + e ^{x} )}$
$Q (x) = \frac{x ^{2}}{x ^{2} + e ^{x}}$

The Integration Trick for $P (x)$

To find IF, we need $\int P (x) d x = \int \frac{e ^{x} ( x - 2 )}{x ( x ^{2} + e ^{x} )} d x$
Divide numerator and denominator by $x^{3}$ :
$\int \frac{\frac{e ^{x}}{x ^{2}} - \frac{2 e ^{x}}{x ^{3}}}{1 + \frac{e ^{x}}{x ^{2}}} d x$

Calculating the Integrating Factor (IF)

Let $t = 1 + \frac{e ^{x}}{x ^{2}} \Rightarrow d t = (\frac{e ^{x}}{x ^{2}} - \frac{2 e ^{x}}{x ^{3}}) d x$
$\int P (x) d x = ln (1 + \frac{e ^{x}}{x ^{2}})$
$I F = e^{l n (1 + \frac{e ^{x}}{x ^{2}})} = 1 + \frac{e ^{x}}{x ^{2}} = \frac{x ^{2} + e ^{x}}{x ^{2}}$

General Solution Setup

General Solution: $y \cdot I F = \int Q (x) \cdot I F d x$
Substitute values: $y (\frac{x ^{2} + e ^{x}}{x ^{2}}) = \int (\frac{x ^{2}}{x ^{2} + e ^{x}}) \cdot (\frac{x ^{2} + e ^{x}}{x ^{2}}) d x$

Integrating the Right Hand Side

$y (1 + \frac{e ^{x}}{x ^{2}}) = \int 1 d x$
$y (1 + \frac{e ^{x}}{x ^{2}}) = x + C$

Using the Initial Condition $(1, 0)$

At $(1, 0)$ , $x = 1$ and $y = 0$
$0 (1 + \frac{e ^{1}}{1 ^{2}}) = 1 + C$
$0 = 1 + C \Rightarrow C = - 1$

The Particular Solution

Particular Solution: $y (1 + \frac{e ^{x}}{x ^{2}}) = x - 1$
$y = \frac{x - 1}{1 + \frac{e ^{x}}{x ^{2}}} = \frac{x ^{2} ( x - 1 )}{x ^{2} + e ^{x}}$

Finding $y (2)$

Substitute $x = 2$ :
$y (2) = \frac{2 ^{2} ( 2 - 1 )}{2 ^{2} + e ^{2}}$
$y (2) = \frac{4 ( 1 )}{4 + e ^{2}} = \frac{4}{4 + e ^{2}}$

Final Conclusion

Key Takeaway: Always look for patterns in $P (x)$ that might simplify the Integrating Factor.
Final Answer: $y (2) = \frac{4}{4 + e ^{2}}$
Next Challenge: Try solving the same DE with a different initial condition, say $y (1) = 1$ .

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

We begin with the given differential equation:

x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0

To organize this, we divide the entire equation by

d x

to obtain:

x (x^{2} + e^{x}) \frac{d y}{d x} + e^{x} (x - 2) y - x^{3} = 0

By shifting the

x^{3}

term to the right and dividing by the coefficient of

\frac{d y}{d x}

, we reveal the standard linear form

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

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

The Integration Trick

To solve this, we calculate the Integrating Factor,

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

. Our

P (x)

\frac{e ^{x} ( x - 2 )}{x ( x ^{2} + e ^{x} )}

To integrate this, we divide the numerator and denominator by

x^{3}

\int \frac{\frac{e ^{x}}{x ^{2}} - \frac{2 e ^{x}}{x ^{3}}}{1 + \frac{e ^{x}}{x ^{2}}} d x

Notice that the numerator is the derivative of the denominator. By substituting

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

, we find

d t = (\frac{e ^{x}}{x ^{2}} - \frac{2 e ^{x}}{x ^{3}}) d x

. The integral simplifies to:

\int \frac{1}{t} d t = ln (t) = ln (1 + \frac{e ^{x}}{x ^{2}})

Thus, our Integrating Factor is:

I F = e^{l n (1 + \frac{e ^{x}}{x ^{2}})} = 1 + \frac{e ^{x}}{x ^{2}}

The Magic of Cancellation

With our

I F

determined, we set up the general solution

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

. Substituting our values yields:

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

The terms cancel elegantly, leaving us with:

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

Final Calculation

We use the initial condition

(1, 0)

to determine the constant

C

. Plugging in

x = 1

and

y = 0

0 (1 + e) = 1 + C \Rightarrow C = - 1

Our particular solution is

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

. Simplifying this, we obtain:

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

Finally, substituting

x = 2

to find

y (2)

y (2) = \frac{2 ^{2} ( 2 - 1 )}{2 ^{2} + e ^{2}} = \frac{4}{4 + e ^{2}}

Common Traps & FAQs

Wait, why can't we just use variable separation here?

Great question! Variable separation works when you can write the equation as

f (y) d y = g (x) d x

. Here, the

y

term is multiplied by a function of

x

and added to another function of

x

, making it impossible to isolate

y

and

x

on opposite sides. This is why we must use the linear differential equation method.

How did you know to divide the numerator and denominator by

x^{3}

in the integral?

That is a classic pattern recognition skill in JEE. When you see a complex fraction involving

e^{x}

and powers of

x

, look at the denominator. The derivative of

\frac{e ^{x}}{x ^{2}}

\frac{x ^{2} e ^{x} - 2 x e ^{x}}{x ^{4}} = \frac{x e ^{x} - 2 e ^{x}}{x ^{3}}

. By dividing by

x^{3}

, we force this derivative to appear in the numerator. It is a powerful technique to keep in your toolkit!

Why do we add the constant

C

only on the right side?

In the general solution

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

, the constant

C

is added after performing the integration on the right side. It represents the family of curves that satisfy the differential equation. We then use the initial condition

(1, 0)

to find the specific value of

C

for our unique curve.

Animated Solution for Mathematics - Differential Equations: 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 :

Select Answer:

Visualized Solution

The Given Differential Equation

Rearranging to Standard Linear Form

Identifying $P (x)$ and $Q (x)$

The Integration Trick for $P (x)$

Calculating the Integrating Factor (IF)

General Solution Setup

Integrating the Right Hand Side

Using the Initial Condition $(1, 0)$

The Particular Solution

Finding $y (2)$

Final Conclusion

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Integration Trick

The Magic of Cancellation

Final Calculation

Similar Questions

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 $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to

Let $y (x)$ 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 $y = f (x)$ 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 the solution of the differential equation $\frac{d y}{d x} + e^{x} (x^{2} - 2) y = (x^{2} - 2 x) (x^{2} - 2) e^{2 x}$ satisfies $y (0) = 0$ , then the value of $y (2)$ is _______.

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

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

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 :

If $x = x (y)$ is the solution of the differential equation $y \frac{d x}{d y} = 2 x + y^{3} (y + 1) e^{y}, x (1) = 0$ ; then $x (e)$ is equal to :

Select Answer:

Visualized Solution

The Given Differential Equation

Rearranging to Standard Linear Form

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

The Integration Trick for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x)

Calculating the Integrating Factor (IF)

General Solution Setup

Integrating the Right Hand Side

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

The Particular Solution

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

Final Conclusion

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Integration Trick

The Magic of Cancellation

Final Calculation

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 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 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(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=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 the solution of the differential equation dxdy​+ex(x2−2)y=(x2−2x)(x2−2)e2x satisfies y(0)=0, then the value of y(2) 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 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; } Let y=y(x) be the solution of the differential equation x3dy+(xy−1)dx=0,x>0,y(21​)=3−e. 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:

.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​+2y=f(x), where f(x)={1,0,​x∈[0,1]otherwise​. If y(0)=0, then y(3/2) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=x(y) is the solution of the differential equation ydydx​=2x+y3(y+1)ey,x(1)=0; then x(e) is equal to :

Identifying $P (x)$ and $Q (x)$

The Integration Trick for $P (x)$

Using the Initial Condition $(1, 0)$

Finding $y (2)$

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 $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to

Let $y (x)$ 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 $y = f (x)$ 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 the solution of the differential equation $\frac{d y}{d x} + e^{x} (x^{2} - 2) y = (x^{2} - 2 x) (x^{2} - 2) e^{2 x}$ satisfies $y (0) = 0$ , then the value of $y (2)$ is _______.

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

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

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 :

If $x = x (y)$ is the solution of the differential equation $y \frac{d x}{d y} = 2 x + y^{3} (y + 1) e^{y}, x (1) = 0$ ; then $x (e)$ is equal to :