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Animated Solution for Mathematics - Differential Equations: If $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ and $y (0) = 1$ , then $y (\frac{π}{2})$ is equal to:

Q: Why do we take the absolute value inside the natural log?

In calculus, the integral of $\frac{1}{u} du$ is defined as $\ln|u| + C$. We include the absolute value bars because the natural logarithm function is only defined for positive real numbers. Even if the expression inside might be positive in the specific domain of the problem, it is a rigorous mathematical habit to include the absolute value to ensure the domain of the logarithm is respected.

Select Answer:

Progressive Hints

Hint 1

The differential equation

(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0

is a classic example of a separable equation. Before you start, visualize the variables

x

and

y

as two distinct entities that need to be isolated on opposite sides of the equality sign. Can you move the terms to group all

y

-related components with

d y

and all

x

-related components with

d x

Hint 2

Once you have separated the variables, you will be left with an integral of the form

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

on one side and

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

on the other. Notice the relationship between the numerator and the denominator on the right-hand side. Does the numerator look like the derivative of the denominator? This is a perfect setup for a simple substitution.

Hint 3

After integrating, you will obtain an expression involving natural logarithms. Remember that

ln A + ln B = ln (A B)

. Use this property to combine the logarithmic terms into a single algebraic equation. Don't forget to use the initial condition

y (0) = 1

to determine the constant of integration

C

before you attempt to find

y (\frac{π}{2})

Visualized Solution

Analyze the Differential Equation

Given equation: $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$
Initial condition: $y (0) = 1$
Goal: Find $y (\frac{π}{2})$
Method: Variable Separable Method

Transpose the Terms

Subtract $(y + 1) cos x$ from both sides:
$(2 + sin x) \frac{d y}{d x} = - (y + 1) cos x$

Separate the Variables

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

Set up the Integrals

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

Integrate the Left Side

LHS Integration:
$\int \frac{1}{y + 1} d y = ln ∣ y + 1∣$

Integrate the Right Side

RHS Integration using substitution $u = 2 + sin x, d u = cos x d x$ :
$- \int \frac{c o s x}{2 + s i n x} d x = - ln ∣2 + sin x ∣$

Combine the Results

Combine results and add constant of integration $ln C$ :
$ln ∣ y + 1∣ = - ln ∣2 + sin x ∣ + ln C$
$ln ∣ y + 1∣ + ln ∣2 + sin x ∣ = ln C$

Simplify the Equation

Using property $ln A + ln B = ln (A B)$ :
$ln ∣ (y + 1) (2 + sin x) ∣ = ln C$
Remove logs: $(y + 1) (2 + sin x) = C$

Apply Initial Condition $y (0) = 1$

Substitute $x = 0$ and $y = 1$ to find $C$ :
$(1 + 1) (2 + sin 0) = C$
$2 \times (2 + 0) = C$
$C = 4$

Formulate Particular Solution

The particular solution is:
$(y + 1) (2 + sin x) = 4$

Substitute $x = \frac{π}{2}$

Substitute $x = \frac{π}{2}$ into the particular solution:
$(y + 1) (2 + sin \frac{π}{2}) = 4$
Since $sin \frac{π}{2} = 1$ :
$(y + 1) (2 + 1) = 4$

Solve for $y$

Simplify and solve for $y$ :
$3 (y + 1) = 4$
$y + 1 = \frac{4}{3}$
$y = \frac{4}{3} - 1 = \frac{1}{3}$

Conclusion & Key Takeaways

Final Answer: $y (\frac{π}{2}) = \frac{1}{3}$
Key Takeaway 1: Always check if variables can be separated first.
Key Takeaway 2: Use initial conditions to find the particular solution constant $C$ .
Key Takeaway 3: Logarithmic properties like $ln A + ln B = ln (A B)$ are essential for simplifying DE solutions.

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Analyzing the Setup

We are tasked with solving the differential equation:

(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0

subject to the initial condition

y (0) = 1

. This equation represents a relationship between variables

x

and

y

that can be untangled through the method of separation of variables.

The Great Separation

To begin, we isolate the terms involving

y

and

x

on opposite sides of the equation. Subtracting

(y + 1) cos x

from both sides yields:

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

Dividing both sides by

(y + 1) (2 + sin x)

and multiplying by

d x

, we obtain the separated form:

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

The Art of Integration

With the variables separated, we integrate both sides:

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

The left side integrates directly 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 transforms the integral into:

ln ∣ y + 1∣ = - ln ∣2 + sin x ∣ + C

The Constant of Integration

To simplify the expression, we represent the constant

C

ln C

. Rearranging the equation gives:

ln ∣ y + 1∣ + ln ∣2 + sin x ∣ = ln C

Using the logarithmic property

ln A + ln B = ln (A B)

, we combine the terms:

ln ∣ (y + 1) (2 + sin x) ∣ = ln C

Exponentiating both sides removes the logarithms, resulting in the general solution:

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

The Final Reveal

We apply the initial condition

y (0) = 1

to determine the value of

C

(1 + 1) (2 + sin 0) = C

Since

sin 0 = 0

, we find

2 \times 2 = C

, which means

C = 4

. The particular solution is therefore:

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

To find

y

x = \frac{π}{2}

, we substitute the value into our particular solution:

(y + 1) (2 + sin \frac{π}{2}) = 4

Given

sin \frac{π}{2} = 1

, the equation simplifies to:

3 (y + 1) = 4

y + 1 = \frac{4}{3}

y = \frac{1}{3}

The final value is $y = \frac{1}{3}$ .

Common Traps & FAQs

Wait, why do we use

ln C

instead of just

C

as the constant of integration?

That is a fantastic question! While you could technically use just

C

, using

ln C

is a strategic choice. Since our integration results in logarithmic terms on both sides, keeping the constant in the form of

ln C

allows us to use the property

ln A + ln B = ln (A B)

to eliminate the logarithms entirely. It transforms a complex logarithmic equation into a clean, simple algebraic one, saving you from potential errors later on.

Can I use the product rule in reverse to solve this?

You absolutely can! If you look closely at the original equation

(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0

, you might notice it resembles the derivative of a product:

\frac{d}{d x} [(y + 1) (2 + sin x)] = 0

. This is the 'hidden' beauty of the problem. Integrating both sides directly gives

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

. Both the separation of variables method and the product rule observation lead to the same elegant result.

Why do we take the absolute value inside the natural log?

In calculus, the integral of

\frac{1}{u} d u

is defined as

ln ∣ u ∣ + C

. We include the absolute value bars because the natural logarithm function is only defined for positive real numbers. Even if the expression inside might be positive in the specific domain of the problem, it is a rigorous mathematical habit to include the absolute value to ensure the domain of the logarithm is respected.

Animated Solution for Mathematics - Differential Equations: If $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ and $y (0) = 1$ , then $y (\frac{π}{2})$ is equal to:

Select Answer:

Visualized Solution

Analyze the Differential Equation

Transpose the Terms

Separate the Variables

Set up the Integrals

Integrate the Left Side

Integrate the Right Side

Combine the Results

Simplify the Equation

Apply Initial Condition $y (0) = 1$

Formulate Particular Solution

Substitute $x = \frac{π}{2}$

Solve for $y$

Conclusion & Key Takeaways

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Great Separation

The Art of Integration

The Constant of Integration

The Final Reveal

Similar Questions

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

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:

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If $\frac{d y}{d x} + \frac{2 ^{x - y} ( 2 ^{y} - 1 )}{2 ^{x} - 1} = 0, x, y > 0, y (1) = 1$ , then $y (2)$ is equal to :

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

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_________

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 $y = y (x)$ be the solution of the differential equation $sec x d y + {2 (1 - x) tan x + x (2 - x)} d x = 0$ such that $y (0) = 2$ . Then $y (2)$ is equal to :

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 :

Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (2+sinx)dxdy​+(y+1)cosx=0 and y(0)=1, then y(2π​) is equal to:

Select Answer:

Visualized Solution

Analyze the Differential Equation

Transpose the Terms

Separate the Variables

Set up the Integrals

Integrate the Left Side

Integrate the Right Side

Combine the Results

Simplify the Equation

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

Formulate Particular Solution

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2π​

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

Conclusion & Key Takeaways

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Great Separation

The Art of Integration

The Constant of Integration

The Final Reveal

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​+2x−12x−y(2y−1)​=0,x,y>0,y(1)=1, then y(2) 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),y∈[0,π/2) is the solution of the differential equation secydxdy​−sin(x+y)−sin(x−y)=0, with y(0)=0, then 5y′(2π​) is equal to

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Animated Solution for Mathematics - Differential Equations: If $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ and $y (0) = 1$ , then $y (\frac{π}{2})$ is equal to:

Apply Initial Condition $y (0) = 1$

Substitute $x = \frac{π}{2}$

Solve for $y$

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

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:

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 $\frac{d y}{d x} + \frac{2 ^{x - y} ( 2 ^{y} - 1 )}{2 ^{x} - 1} = 0, x, y > 0, y (1) = 1$ , then $y (2)$ is equal to :

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

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_________

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 $y = y (x)$ be the solution of the differential equation $sec x d y + {2 (1 - x) tan x + x (2 - x)} d x = 0$ such that $y (0) = 2$ . Then $y (2)$ is equal to :