Let a curve y=f(x) pass through the points (0,5) and (log ₑ 2, k). If the curve satisfies the differential equation 2(3+y) e²ˣ d x-(7+e²ˣ) d y=0, then k is equal to

Solution

Maths sir · 4 steps

Put all the y terms with dy and all the x terms with dx, and both sides integrate to a log.

1. Separate variables. Rewrite the equation as (7+e²ˣ) dy = 2(3+y)e²ˣ dx. Divide both sides by (3+y)(7+e²ˣ) so that each side holds only one variable.

   ∫ dy/(3+y) = ∫ 2e²ˣ/(7+e²ˣ) dx

2. Integrate. The numerator 2e²ˣ on the right is the derivative of 7+e²ˣ, so both sides are logs: ln(3+y) = ln(7+e²ˣ) + ln C. This gives 3+y = C(7+e²ˣ).
3. Use the point. At (0,5) we get 3+5 = C(7+1), so 8 = 8C and C = 1. The curve is y = 4 + e²ˣ.
4. So. At x = logₑ 2, e²ˣ = eˡⁿ⁴ = 4, so k = 4 + 4 = 8, which is option (b).

The idea to keep. After separating, look for a numerator that is the derivative of its denominator. That integral is a log, and writing the constant as ln C keeps the algebra clean.

Answer from the NTA final answer key. Solution by PrepWiser.