JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let a curve $y = f (x)$ pass through the points $(0, 5)$ and $(lo g_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to

Q: I got $C = 0$, but I feel like I made a mistake. Is it possible for the constant to be zero?

Absolutely! It is perfectly fine for $C$ to be zero. In fact, it happens quite often in these problems. When you substitute $x=0$ and $y=5$, you get $\ln(8) = \ln(8) + C$. The only way this equation holds true is if $C = 0$. Do not let the simplicity of the result make you doubt your math; trust the process!

Q: How do I handle the term $e^{2\ln 2}$ when calculating $k$?

Remember your logarithm rules! The coefficient $2$ in front of the $\ln 2$ can be moved to the exponent of the $2$, so $2\ln 2$ becomes $\ln(2^2)$, which is $\ln 4$. Then, $e^{\ln 4}$ is simply $4$ because the exponential and natural logarithm functions are inverses of each other. It is a beautiful simplification!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Look at the differential equation

2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0

. This is a classic separable variable problem. Your first task is to rearrange the terms so that all

x

-dependent terms are on one side with

d x

and all

y

-dependent terms are on the other with

d y

. Think of it as organizing your workspace before you start the real work.

Hint 2

Hint 2: The Arsenal. Once you have separated the variables, you will have an integral of the form

\int \frac{2 e ^{2 x}}{7 + e ^{2 x}} d x

on one side and

\int \frac{1}{y + 3} d y

on the other. For the left side, notice that the numerator is the derivative of the denominator. What substitution should you use to make this integral trivial?

Hint 3

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

ln (7 + e^{2 x}) = ln (y + 3) + C

. Use the initial condition

(0, 5)

to find the value of the constant

C

. Remember that

e^{0} = 1

. Once you have

C

, you can solve for

y

in terms of

x

and finally substitute

x = ln 2

to find

k

Visualized Solution

The Differential Equation

Given: $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$
The curve passes through $(0, 5)$ and $(lo g_{e} 2, k)$ .

Variable Separation

Rearrange to group $x$ and $y$ terms:
$\Rightarrow 2 (3 + y) e^{2 x} d x = (7 + e^{2 x}) d y$
$\Rightarrow \frac{2 e ^{2 x}}{7 + e ^{2 x}} d x = \frac{d y}{y + 3}$

Setting up the Integrals

Integrate both sides:
$\int \frac{2 e ^{2 x}}{7 + e ^{2 x}} d x = \int \frac{1}{y + 3} d y$

Integrating the Left Hand Side

Let $u = 7 + e^{2 x}$ , then $d u = 2 e^{2 x} d x$
$\int \frac{1}{u} d u = ln ∣7 + e^{2 x} ∣$

Integrating the Right Hand Side

$\int \frac{1}{y + 3} d y = ln ∣ y + 3∣$

General Solution

Equating both sides and adding a constant $C$ :
$ln (7 + e^{2 x}) = ln (y + 3) + C$

Applying the Initial Condition

Substitute the point $(0, 5)$ to find $C$ :
$ln (7 + e^{0}) = ln (5 + 3) + C$

Solving for $C$

$ln (7 + 1) = ln (8) + C$
$ln (8) = ln (8) + C$
$\Rightarrow C = 0$

The Particular Solution

Substitute $C = 0$ back:
$ln (7 + e^{2 x}) = ln (y + 3)$
$\Rightarrow 7 + e^{2 x} = y + 3$
$\Rightarrow y = 4 + e^{2 x}$

Finding the Target Point

The curve passes through $(lo g_{e} 2, k)$ .
Substitute $x = lo g_{e} 2$ and $y = k$ :
$k = 4 + e^{2 l o g_{e} 2}$

Evaluating $k$

Using logarithm property: $e^{2 l o g_{e} 2} = e^{l o g_{e} (2^{2})} = e^{l o g_{e} 4}$
$\Rightarrow k = 4 + 4$
$\Rightarrow k = 8$

Final Conclusion

The value of $k$ is $8$ .
The point is $(lo g_{e} 2, 8)$ .

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

Imagine you are standing before a complex, winding path on a graph defined by the differential equation:

2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0

We are given that this curve passes through two specific points:

(0, 5)

and

(ln 2, k)

. Our mission is to determine the value of

k

The Great Separation

The first step is to bring order to the chaos by separating the variables. We rearrange the equation to group all

x

terms with

d x

and all

y

terms with

d y

2 (3 + y) e^{2 x} d x = (7 + e^{2 x}) d y

Dividing both sides to isolate the variables, we obtain:

\frac{2 e ^{2 x}}{7 + e ^{2 x}} d x = \frac{d y}{y + 3}

The Art of Integration

Now, we apply the integral operator to both sides of the equation:

\int \frac{2 e ^{2 x}}{7 + e ^{2 x}} d x = \int \frac{1}{y + 3} d y

The right side integrates to

ln ∣ y + 3∣

. For the left side, we use the substitution

u = 7 + e^{2 x}

, which implies

d u = 2 e^{2 x} d x

This transforms the integral into

\int \frac{1}{u} d u

, resulting in

ln ∣ u ∣

. Substituting back, we arrive at the general solution:

ln (7 + e^{2 x}) = ln (y + 3) + C

Finding the Anchor

To find the specific curve, we use the anchor point

(0, 5)

. Substituting

x = 0

and

y = 5

into our general solution:

ln (7 + e^{0}) = ln (5 + 3) + C

Since

e^{0} = 1

, this simplifies to

ln (8) = ln (8) + C

, which yields

C = 0

. Consequently, the specific equation of the curve is:

ln (7 + e^{2 x}) = ln (y + 3)

By removing the natural logarithms, we simplify this to

7 + e^{2 x} = y + 3

, or:

y = 4 + e^{2 x}

The Final Destination

We now find

k

at the point

(ln 2, k)

by substituting

x = ln 2

into our specific curve equation:

k = 4 + e^{2 l n 2}

Using the logarithmic property

2 ln 2 = ln (2^{2}) = ln 4

, we get:

k = 4 + e^{l n 4}

Since

e

and

ln

are inverse functions,

e^{l n 4} = 4

. Therefore, the final value is:

$k = 4 + 4 = 8$

Common Traps & FAQs

Wait, why do we need to add the constant

C

after integrating? Can't we just ignore it?

That is a great question! In differential equations, the constant

C

represents the entire family of curves that satisfy the equation. Without it, you are only looking at one specific curve. Since we are given a point

(0, 5)

that the curve must pass through, we need that constant to 'pin' our curve to that specific location in the coordinate plane. It is the difference between a general solution and the specific solution we need.

I got

C = 0

, but I feel like I made a mistake. Is it possible for the constant to be zero?

Absolutely! It is perfectly fine for

C

to be zero. In fact, it happens quite often in these problems. When you substitute

x = 0

and

y = 5

, you get

ln (8) = ln (8) + C

. The only way this equation holds true is if

C = 0

. Do not let the simplicity of the result make you doubt your math; trust the process!

How do I handle the term

e^{2 l n 2}

when calculating

k

Remember your logarithm rules! The coefficient

2

in front of the

ln 2

can be moved to the exponent of the

2

, so

2 ln 2

becomes

ln (2^{2})

, which is

ln 4

. Then,

e^{l n 4}

is simply

4

because the exponential and natural logarithm functions are inverses of each other. It is a beautiful simplification!

Animated Solution for Mathematics - Differential Equations: Let a curve $y = f (x)$ pass through the points $(0, 5)$ and $(lo g_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to

Select Answer:

Visualized Solution

The Differential Equation

Variable Separation

Setting up the Integrals

Integrating the Left Hand Side

Integrating the Right Hand Side

General Solution

Applying the Initial Condition

Solving for $C$

The Particular Solution

Finding the Target Point

Evaluating $k$

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Great Separation

The Art of Integration

Finding the Anchor

The Final Destination

Similar Questions

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 $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to

Let $y = y (x)$ be solution of the differential equation $lo g_{e} (\frac{d y}{d x}) = 3 x + 4 y$ , with $y (0) = 0$ . If $y (- \frac{2}{3} lo g_{e} 2) = α lo g_{e} 2$ , then the value of $α$ 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

If the solution of the differential equation $(2 x + 3 y - 2) d x + (4 x + 6 y - 7) d y = 0, y (0) = 3$ , is $α x + β y + 3 lo g_{e} ∣2 x + 3 y - γ ∣ = 6$ , then $α + 2 β + 3 γ$ 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 a curve $y = f (x)$ pass through the point $(2, (lo g_{e} 2)^{2})$ and have slope $\frac{2 y}{x l o g _{e} x}$ for all positive real value of $x$ . Then the value of $f (e)$ is equal to ___

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

If $y (x)$ is the solution of the differential equation $x d y - (y^{2} - 4 y) d x = 0$ for $x > 0$ , $y (1) = 2$ , and the slope of the curve $y = y (x)$ is never zero, then the value of $10 y (2)$ is ______.

Select Answer:

Visualized Solution

The Differential Equation

Variable Separation

Setting up the Integrals

Integrating the Left Hand Side

Integrating the Right Hand Side

General Solution

Applying the Initial Condition

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

The Particular Solution

Finding the Target Point

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

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Great Separation

The Art of Integration

Finding the Anchor

The Final Destination

Similar Questions

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.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 dy=eαx+ydx;α∈N. If y(loge​2)=loge​2 and y(0)=loge​(1/2), then the value of α 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 solution of the differential equation loge​(dxdy​)=3x+4y, with y(0)=0. If y(−32​loge​2)=αloge​2, then the value of α 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; } If the solution of the differential equation (2x+3y−2)dx+(4x+6y−7)dy=0,y(0)=3, is αx+βy+3loge​∣2x+3y−γ∣=6, then α+2β+3γ 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 a curve y=f(x) pass through the point (2,(loge​2)2) and have slope xloge​x2y​ for all positive real value of x. Then the value of f(e) 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, ey(dxdy​−1)=ex such that y(0)=0, then y(1) is equal to

Solving for $C$

Evaluating $k$

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 $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to

Let $y = y (x)$ be solution of the differential equation $lo g_{e} (\frac{d y}{d x}) = 3 x + 4 y$ , with $y (0) = 0$ . If $y (- \frac{2}{3} lo g_{e} 2) = α lo g_{e} 2$ , then the value of $α$ 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

If the solution of the differential equation $(2 x + 3 y - 2) d x + (4 x + 6 y - 7) d y = 0, y (0) = 3$ , is $α x + β y + 3 lo g_{e} ∣2 x + 3 y - γ ∣ = 6$ , then $α + 2 β + 3 γ$ 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 a curve $y = f (x)$ pass through the point $(2, (lo g_{e} 2)^{2})$ and have slope $\frac{2 y}{x l o g _{e} x}$ for all positive real value of $x$ . Then the value of $f (e)$ is equal to ___

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