Exams › JEE Advanced › Maths
Correct answer: 3y + 2 = x²
(x³-xy)dx = (1+x²)dy. Rearranging: dy/dx = (x³-xy)/(1+x²). => dy/dx + xy/(1+x²) = x³/(1+x²). Integrating factor: mu = exp(integral of x/(1+x²) dx) = exp((1/2)*ln(1+x²)) = sqrt(1+x²). Multiplying through: d[y*sqrt(1+x²)]/dx = x³/sqrt(1+x²). RHS: integral of x³/sqrt(1+x²) dx. Let u = 1+x², du = 2x dx, x² = u-1. integral = integral of (u-1)*x/sqrt(u) * du/(2x) = (1/2)*integral(u-1)/sqrt(u) du = (1/2)*integral(sqrt(u) - 1/sqrt(u)) du = (1/2)*(2u^(3/2)/3 - 2u^(1/2)) + C = u^(3/2)/3 - u^(1/2) + C = sqrt(u)*(u/3 - 1) + C = sqrt(1+x²)*((1+x²)/3 - 1) + C = sqrt(1+x²)*(x²-2)/3 + C. So y*sqrt(1+x²) = sqrt(1+x²)*(x²-2)/3 + C. => y = (x²-2)/3 + C/sqrt(1+x²). Apply y(0) = -2/3: -2/3 = (0-2)/3 + C/1 = -2/3 + C => C = 0. Therefore y = (x²-2)/3 => 3y = x²-2 => x² - 3y - 2 = 0... rearranging: x² + (-3y) - 2 = 0. Checking option: x² + 3y + 2 = 0 => 3y = -x²-2 => y = -(x²+2)/3. At x=0: y = -2/3. Check! Check ODE: y = -(x²+2)/3. dy/dx = -2x/3. From ODE: dy/dx = (x³-xy)/(1+x²) = (x³-x*(-(x²+2)/3))/(1+x²) = (x³ + x(x²+2)/3)/(1+x²) = x*(3x² + x²+2)/(3*(1+x²)) = x*(4x²+2)/(3*(1+x²)) = 2x*(2x²+1)/(3*(1+x²)). This should equal -2x/3: 2*(2x²+1)/(3*(1+x²)) = -2/3? That gives 2x²+1 = -(1+x²), 3x² = -2, impossible. Let me redo: y*sqrt(1+x²) = sqrt(1+x²)*(x²-2)/3 + C => y = (x²-2)/3 + C/sqrt(1+x²). At x=0: -2/3 = -2/3 + C => C=0. y=(x²-2)/3. 3y=x²-2 => x²-3y-2=0. Hmm, that doesn't match any option exactly. x²-3y-2=0 => x² = 3y+2 => option 'x² = 3y+2'? Looking at option D: '3y + 2 = x²' which is x² = 3y+2, i.e., 3y+2 = x². Let's verify y=(x²-2)/3 satisfies this: 3*(x²-2)/3 + 2 = x²-2+2 = x². Yes! So option D: 3y+2 = x² is correct. But wait, option A is x²+3y+2=0, which gives y=-(x²+2)/3, giving y(0)=-2/3. Let me verify option A in ODE: y=-(x²+2)/3, dy/dx=-2x/3. RHS = (x³-xy)/(1+x²) = (x³+x(x²+2)/3)/(1+x²) = x*(3x²+x²+2)/(3(1+x²)) = 2x(2x²+1)/(3(1+x²)). At x=0 this gives 0 = dy/dx=0. At x=1: RHS=2*3/6=1, LHS dy/dx=-2/3. Not equal. So option A is wrong. Option D: 3y+2=x², y=(x²-2)/3, dy/dx=2x/3. RHS=(x³-xy)/(1+x²)=(x³-x(x²-2)/3)/(1+x²)=(3x³-x³+2x)/(3(1+x²))=(2x³+2x)/(3(1+x²))=2x(x²+1)/(3(1+x²))=2x/3. Yes! LHS=RHS. Option D is correct.