If the curve y = y(x) is the solution of the differential equation
2(x² + x^(5/4))dy - y(x + x^(1/4))dx = 2x^(9/4)dx, x > 0 which
passes through the point ( 1,1 - 4/3log ₑ2 ), then the value of y(16) is equal to :

Solution

Maths sir · 5 steps

Pull out the common factor (x^(3/4)+1) first, and this turns into a clean linear equation in y.

1. Tidy the coefficients. Factor x² + x^(5/4) = x^(5/4)(x^(3/4)+1) and x + x^(1/4) = x^(1/4)(x^(3/4)+1). Divide the whole equation by 2x^(5/4)(x^(3/4)+1) dx, and the factor (x^(3/4)+1) cancels from the y term.

   dy/dx - y/2x = x/(x^(3/4)+1)

2. Integrating factor. Here P = -1/2x, so IF = e^(-1/2 ln x) = x^(-1/2). The equation becomes y/√x = ∫ x^(1/2)/(x^(3/4)+1) dx.
3. Do the integral. Put x = t⁴, so dx = 4t³ dt and the integral is ∫ 4t⁵/(t³+1) dt = 4∫ (t² - t²/(t³+1))dt = 4/3(t³ - ln(1+t³)). With t³ = x^(3/4), this gives y/√x = 4/3(x^(3/4) - ln(1+x^(3/4))) + C.
4. Find C. At x = 1, y = 1 - 4/3 ln 2. So 1 - 4/3 ln 2 = 4/3(1 - ln 2) + C, which gives C = -1/3.
5. So. At x = 16, √x = 4 and x^(3/4) = 8, so y/4 = 4/3(8 - ln 9) - 1/3 = 31/3 - 8/3 ln 3. Hence y(16) = 4(31/3 - 8/3 ln 3), option (a).

The idea to keep. When a messy DE has fractional powers, factor first and look for a common bracket to cancel. Then for powers like x^(1/4), substitute x = t⁴ to clear the fractional exponents.

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