Let the solution y=y(x) of the differential equation (d y)/(d x)-y=1+4 sin x satisfy y(π)=1. Then y(π/2)+10 is equal to __________.

Solution

Maths sir · 5 steps

Multiply by the integrating factor e⁻ˣ, and the condition at x = π kills the constant.

1. Integrating factor. The equation has the form y' + Py = Q with P = -1. So the integrating factor is e^(∫ -1 dx) = e⁻ˣ, and the left side becomes d/dx(y e⁻ˣ).

   y e⁻ˣ = ∫ e⁻ˣ dx + 4∫ e⁻ˣ sin x dx

2. Integrate each part. ∫ e⁻ˣ dx = -e⁻ˣ. Using ∫ eᵃˣ sin bx dx = eᵃˣ(a sin bx - b cos bx)/(a²+b²) with a = -1, b = 1 gives ∫ e⁻ˣ sin x dx = -e⁻ˣ(sin x + cos x)/2. So y e⁻ˣ = -e⁻ˣ - 2e⁻ˣ(sin x + cos x) + C.
3. General solution. Divide by e⁻ˣ: y = -1 - 2 sin x - 2 cos x + Ceˣ.
4. Use y(π) = 1. At x = π, sin π = 0 and cos π = -1, so 1 = -1 + 2 + Ce^π. This gives C = 0, and y = -1 - 2 sin x - 2 cos x.
5. So. y(π/2) = -1 - 2 - 0 = -3, so y(π/2) + 10 = 7.

The common slip: -3. Stopping at y(π/2) gives -3. The question asks for that value plus 10.

The idea to keep. For y' + Py = Q, multiply by e^(∫ P dx) and integrate. Keep the standard formula for ∫ eᵃˣ sin bx dx ready, because the signs are where marks are lost.

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