JEE Main 2020 - 2 Sep (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation, $\frac{2 + s i n x}{y + 1} \frac{d y}{d x} = - cos x$ , $y > 0$ , $y (0) = 1$ . If $y (π) = a$ and $\frac{d y}{d x}$ at $x = π$ is $b$ , then the ordered pair $(a, b)$ is equal to :

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): This is a classic separable differential equation. Do not let the trigonometric functions intimidate you. Observe the structure of the right-hand side; it is a perfect candidate for a simple substitution. Look for the relationship between the numerator and the denominator.

Hint 2

Hint 2 (The Arsenal): Recall the integration technique where the integral of

\frac{f ^{'} ( x )}{f ( x )} d x

ln ∣ f (x) ∣ + C

. When you separate the variables, you will find that the integral on the right side follows this exact pattern.

Hint 3

Hint 3 (The Execution): Start by separating the variables to get

\int \frac{d y}{y + 1} = - \int \frac{c o s x}{2 + s i n x} d x

. After integrating, use the initial condition

y (0) = 1

to determine the constant of integration

C

. Finally, use the resulting particular solution to find

a = y (π)

and the original differential equation to find

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

x = π

Visualized Solution

Analyze the Differential Equation

Given Differential Equation: $\frac{2 + s i n x}{y + 1} \frac{d y}{d x} = - cos x$
Initial Condition: $y (0) = 1$
Goal: Find the ordered pair $(a, b)$ where $a = y (π)$ and $b = \frac{d y}{d x}$ at $x = π$

Separating the Variables

Rearrange the equation to separate $y$ and $x$ terms:
$\frac{d y}{y + 1} = - \frac{c o s x}{2 + s i n x} d x$

Applying Integration

Integrate both sides of the equation:
$\int \frac{d y}{y + 1} = - \int \frac{c o s x}{2 + s i n x} d x$

Solving the Integrals

Left side integral: $\int \frac{d y}{y + 1} = ln ∣ y + 1∣$
Right side substitution: Let $u = 2 + sin x ⟹ d u = cos x d x$
Right side integral: $- \int \frac{d u}{u} = - ln ∣2 + sin x ∣ + ln C$

The General Solution

Equating both sides: $ln ∣ y + 1∣ = - ln ∣2 + sin x ∣ + ln C$
Rearranging: $ln ∣ y + 1∣ + ln ∣2 + sin x ∣ = ln C$
Using log properties: $ln ∣ (y + 1) (2 + sin x) ∣ = ln C$
General Solution: $(y + 1) (2 + sin x) = C$

Finding the Constant $C$

Apply initial condition $y (0) = 1$ :
$(1 + 1) (2 + sin 0) = C$
$2 (2 + 0) = C ⟹ C = 4$

The Particular Solution

Substitute $C = 4$ back into the general solution:
Particular Solution: $(y + 1) (2 + sin x) = 4$

Finding the Value of $a$

To find $a = y (π)$ , substitute $x = π$ into the particular solution:
$(y (π) + 1) (2 + sin π) = 4$
$(a + 1) (2 + sin π) = 4$

Calculating $a$

Since $sin π = 0$ :
$(a + 1) (2 + 0) = 4$
$2 (a + 1) = 4 ⟹ a + 1 = 2$
$a = 1$

Finding the Derivative $b$

From the original differential equation, isolate $\frac{d y}{d x}$ :
$\frac{d y}{d x} = - \frac{c o s x ( y + 1 )}{2 + s i n x}$
We need to evaluate this at $x = π$ and $y = a = 1$ .

Calculating $b$

Substitute $x = π$ and $y = 1$ :
$b = - \frac{c o s π ( 1 + 1 )}{2 + s i n π}$
$b = - \frac{( - 1 ) ( 2 )}{2 + 0}$
$b = \frac{2}{2} = 1$

Final Result

The value of $a = 1$ .
The value of $b = 1$ .
The ordered pair $(a, b)$ is $(1, 1)$ .
Correct Option: (4)

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

The given differential equation is:

\frac{2 + sin x}{y + 1} \frac{d y}{d x} = - cos x

We are provided with the initial condition

y (0) = 1

. Our objective is to determine the value

a = y (π)

and the slope

b = \frac{d y}{d x}_{x = π}

The Art of Separation

To solve this, we must isolate the variables

x

and

y

. By rearranging the terms, we move all

y

components to the left and all

x

components to the right:

\frac{d y}{y + 1} = - \frac{cos x}{2 + sin x} d x

This separation transforms the complex relationship into two distinct, integrable parts.

The Integration Dance

We now apply the integral operator to both sides of the equation:

\int \frac{d y}{y + 1} = - \int \frac{cos x}{2 + sin x} d x

The left side integrates to

ln ∣ y + 1∣

. For the right side, we use the substitution

u = 2 + sin x

, which implies

d u = cos x d x

This substitution yields:

ln ∣ y + 1∣ = - ln ∣2 + sin x ∣ + ln C

The Particular Path

Using logarithmic properties, we combine the terms:

ln ∣ y + 1∣ + ln ∣2 + sin x ∣ = ln C

ln ∣ (y + 1) (2 + sin x) ∣ = ln C

Exponentiating both sides gives the general solution:

(y + 1) (2 + sin x) = C

Applying the initial condition

y (0) = 1

(1 + 1) (2 + sin 0) = C \Rightarrow 2 (2) = C \Rightarrow C = 4

Thus, the particular solution is

(y + 1) (2 + sin x) = 4

The Final Reveal

To find

a = y (π)

, we substitute

x = π

into the particular solution:

(a + 1) (2 + sin π) = 4

Since

sin π = 0

, we have

2 (a + 1) = 4

, which simplifies to

a + 1 = 2

, or $a = 1$ .

To find the slope

b

, we rearrange the original differential equation:

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

Substituting

x = π

and

y = 1

b = - \frac{cos π ( 1 + 1 )}{2 + sin π} = - \frac{( - 1 ) ( 2 )}{2 + 0} = \frac{2}{2} = 1

The final result is the ordered pair $(a, b) = (1, 1)$ .

Common Traps & FAQs

Why do we use

ln C

as the constant of integration instead of just

C

That is a fantastic question! While you can technically use

C

, using

ln C

is a strategic choice. Since both sides of our equation result in natural logarithms, using

ln C

allows us to combine all terms into a single logarithmic expression. This makes it much easier to 'exponentiate' both sides and eliminate the logs entirely, leading to a cleaner algebraic form for our general solution.

Do I really need to differentiate the particular solution to find the slope

b

Not at all! In fact, that would be the long way around. Remember, the original differential equation

\frac{2 + s i n x}{y + 1} \frac{d y}{d x} = - cos x

already gives you an expression for

\frac{d y}{d x}

. Once you have the particular solution, you know the relationship between

x

and

y

. You can simply plug the values of

x

and

y

at the point of interest directly into the original differential equation to find the slope

b

instantly.

What if the denominator

2 + sin x

becomes zero?

That is a great observation regarding the domain. However, in this problem, we are dealing with the function

sin x

, which oscillates between

- 1

and

1

. Therefore,

2 + sin x

will always be between

1

and

3

. It will never be zero, so our solution is well-defined for all real values of

x

Select Answer:

Visualized Solution

Analyze the Differential Equation

Separating the Variables

Applying Integration

Solving the Integrals

The General Solution

Finding the Constant $C$

The Particular Solution

Finding the Value of $a$

Calculating $a$

Finding the Derivative $b$

Calculating $b$

Final Result

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Integration Dance

The Particular Path

The Final Reveal

Similar Questions

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If the solution $y (x)$ of the given differential equation $(e^{y} + 1) cos x d x + e^{y} sin x d y = 0$ passes through the point $(\frac{π}{2}, 0)$ , then the value of $e^{y (\frac{π}{6})}$ is equal to_________

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Select Answer:

Visualized Solution

Analyze the Differential Equation

Separating the Variables

Applying Integration

Solving the Integrals

The General Solution

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

The Particular Solution

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

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

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

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

Final Result

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Integration Dance

The Particular Path

The Final Reveal

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Finding the Constant $C$

Finding the Value of $a$

Calculating $a$

Finding the Derivative $b$

Calculating $b$

If $y = y (x)$ and $\frac{2 + s i n x}{y + 1} (\frac{d y}{d x}) = - cos x, y (0) = 1$ , then $y (\frac{π}{2})$ equals

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If $y = y (x), y \in [0, π /2)$ is the solution of the differential equation $sec y \frac{d y}{d x} - sin (x + y) - sin (x - y) = 0$ , with $y (0) = 0$ , then $5 y^{'} (\frac{π}{2})$ 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

Let $y = y (x)$ be a solution of the differential equation, $1 - x^{2} \frac{d y}{d x} + 1 - y^{2} = 0, ∣ x ∣ < 1$ . If $y (\frac{1}{2}) = \frac{3}{2}$ , then $y (\frac{- 1}{2})$ is equal to :

If the solution $y = y (x)$ of the differential equation $(x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 2) d y - (2 x^{2} + 2 x + 3) d x = 0$ satisfies $y (- 1) = - \frac{π}{4}$ , then $y (0)$ is equal to :

If the solution $y (x)$ of the given differential equation $(e^{y} + 1) cos x d x + e^{y} sin x d y = 0$ passes through the point $(\frac{π}{2}, 0)$ , then the value of $e^{y (\frac{π}{6})}$ is equal to_________

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