Question

Let a curve \( y = f(x) \) pass through the points \( (0,5) \) and \( (\log 2, k) \). If the curve satisfies the differential equation: \[ 2(3+y)e^{2x}dx - (7+e^{2x})dy = 0, \] then \( k \) is equal to:

• A. \( 16 \)
• B. \( 8 \)
• C. \( 32 \)
• D. \( 4 \)

Hint:

When solving a differential equation, always check whether it's separable or requires an integrating factor. Substituting boundary conditions correctly helps determine constants.

Answer 1

The Correct Option is B

To solve the given problem, we need to determine the value of \( k \) for which the curve passes through the points \((0, 5)\) and \((\log 2, k)\), and satisfies the differential equation:

\[2(3+y)e^{2x}dx - (7+e^{2x})dy = 0.\]

Let's solve this differential equation step by step: 

1. Rewrite the given differential equation to separate variables: 

\[2(3+y)e^{2x} \, dx = (7+e^{2x}) \, dy.\]

1. Separate the variables: 

\[\frac{dy}{2(3+y)} = \frac{e^{2x} \, dx}{7+e^{2x}}.\]

1. Integrate both sides:

Left side integration: 

\[\int \frac{dy}{2(3+y)} = \frac{1}{2} \int \frac{1}{3+y} \, dy = \frac{1}{2} \ln|3+y|.\]

Right side integration: 

\[\int \frac{e^{2x}}{7+e^{2x}} \, dx.\]

For integration purpose, let \( t = 7 + e^{2x} \), then \( dt = 2e^{2x} \, dx \) or \( \frac{1}{2}dt = e^{2x} \, dx \). Thus: 

\[\frac{1}{2} \int \frac{1}{t} \, dt = \frac{1}{2} \ln|t| = \frac{1}{2} \ln|7+e^{2x}|.\]

1. After integration, we have: 

\[\frac{1}{2} \ln|3+y| = \frac{1}{2} \ln|7+e^{2x}| + C.\]

1. Simplify the equation by removing logs with base \(e\): 

\[\ln|3+y| = \ln|7+e^{2x}| + C.\]

1. Remove logarithms by exponentiation: 

\[|3+y| = A|7+e^{2x}|,\]

1. where \(A = e^C\).
2. Use initial condition \((0, 5)\): 

\[|3+5| = A|7+e^{0}| \implies 8 = A(7+1) \rightarrow A = 1.\]

1. Thus, the equation becomes: 

\[3+y = 7+e^{2x}.\]

1. The function \( y \) is: 

\[y = 4 + e^{2x}.\]

1. Now, substitute \( x = \log 2 \) to find \( k \): 

\[y = 4 + e^{2\log 2} = 4 + (2^2) = 4 + 4 = 8.\]

Hence, the value of \( k \) is 8.

Answer 2

Step 1: Expressing the differential equation. Rewriting the given equation: \[ \frac{dy}{dx} = \frac{2(3+y)e^{2x}}{7+e^{2x}}. \] Separating variables: \[ \frac{dy}{(3+y)} = \frac{2e^{2x}dx}{7+e^{2x}}. \] Step 2: Finding the integrating factor (I.F.). \[ I.F. = e^{-\int \frac{2e^{2x}dx}{7+e^{2x}}}. \] Using substitution, \[ I.F. = \frac{1}{7 + e^{2x}}. \] Step 3: Solving for \( y \). Multiplying by the integrating factor: \[ y \cdot \frac{1}{7+e^{2x}} = \int \frac{6e^{2x}dx}{(7+e^{2x})^2}. \] Integrating, we get: \[ \frac{y}{7+e^{2x}} = \frac{-3}{7+e^{2x}} + C. \] Step 4: Applying the initial condition \( (0,5) \). \[ \frac{5}{8} = \frac{-3}{8} + C. \] Solving for \( C \), \[ C = 1. \] Step 5: Finding \( k \) at \( x = \log 2 \). \[ y = -3 + 7 + e^{2x} = e^{2x} + 4. \] \[ k = e^{2 \log 2} + 4 = 4 + 4 = 8. \]