JEE Main 2023 · 30 January · evening · Single correct · +4 / −1
The solution of the differential equation
(d y)/(d x)=-((x²+3 y²)/(3 x²+y²)), y(1)=0 is :
- (a) log ₑ|x+y|+(x y)/(x+y)²=0
- (b) log ₑ|x+y|-(x y)/(x+y)²=0
- (c) log ₑ|x+y|+(2 x y)/(x+y)²=0
- (d) log ₑ|x+y|-(2 x y)/(x+y)²=0
Answer and solution
Answer (c) log ₑ|x+y|+(2 x y)/(x+y)²=0
Solution
Maths sir · 4 steps
Put y = vx, then shift to t = 1 + v, and the messy fraction splits into three easy powers of t.
- Homogeneous substitution. The right side depends only on v = y/x, so the equation is homogeneous. With dy/dx = v + xdv/dx, you get xdv/dx = -(1+3v²)/(3+v²) - v = -(1 + 3v + 3v² + v³)/(3+v²) = -(1+v)³/(3+v²).
- Separate and shift. This gives (3+v²)/(1+v)³ dv = -dx/x. Put t = 1+v, so 3 + v² = t² - 2t + 4 and the left side becomes (1/t - 2/t² + 4/t³)dt.
ln|t| + 2/t - 2/t² = -ln|x| + C
- Undo the substitution. ln|1+v| + ln|x| = ln|x+y|, and 2/t - 2/t² = 2(t-1)/t² = 2v/(1+v)² = 2xy/(x+y)². So ln|x+y| + 2xy/(x+y)² = C.
- So. At x = 1, y = 0 both terms are 0, so C = 0. The solution is logₑ|x+y| + 2xy/(x+y)² = 0, which is option (c).
The idea to keep. When a homogeneous equation leaves (1+v)ⁿ in the denominator, put t = 1+v so the fraction splits into plain powers of t. Every option passes through (1,0), so the integration itself must decide.
Answer from the NTA final answer key. Solution by PrepWiser.