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 must determine the value of \( k \) for which the curve passes through the points \((0, 5)\) and \((\log 2, k)\) and satisfies the differential equation:

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

The differential equation is solved as follows:

1. Rewrite the 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:

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

Using the substitution \( t = 7 + e^{2x} \), where \( dt = 2e^{2x} \, dx \) or \( \frac{1}{2}dt = e^{2x} \, dx \):

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

1. After integration:

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

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. Apply the initial condition \((0, 5)\):

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

1. The resulting equation is:

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

1. The function \( y \) is:

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

1. Substitute \( x = \log 2 \) to find \( k \):

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

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