JEE Main 2022 · 25 July · morning · Single correct · +4 / −1
The general solution of the differential equation (x-y²) d x+y(5 x+y²) d y=0 is :
- (a) (y²+x)⁴=C|(y²+2 x)³|
- (b) (y²+2 x)⁴=C|(y²+x)³|
- (c) |(y²+x)³|=C(2 y²+x)⁴
- (d) |(y²+2 x)³|=C(2 y²+x)⁴
Answer and solution
Answer (a) (y²+x)⁴=C|(y²+2 x)³|
Solution
Maths sir · 4 steps
Put t = y² first: the equation becomes homogeneous in x and t, and then t = vx finishes it.
- Substitute y squared. Let t = y², so y dy = 1/2dt. The equation becomes (x - t) dx + 1/2(5x + t) dt = 0, that is dt/dx = 2(t - x)/(5x + t). Every term has degree 1 in x and t, so it is homogeneous.
- Put t = vx. With dt/dx = v + xdv/dx, we get xdv/dx = (2v - 2)/(5 + v) - v = -(v² + 3v + 2)/(v + 5) = -(v+1)(v+2)/(v+5). Now separate the variables.
∫ (4/(v+1) - 3/(v+2)) dv = -∫ dx/x
- Integrate. The partial fractions come from A = (-1+5)/(-1+2) = 4 and B = (-2+5)/(-2+1) = -3. Integrating gives 4 ln|v+1| - 3 ln|v+2| + ln|x| = c.
- So. Put back v + 1 = (t + x)/x and v + 2 = (t + 2x)/x. The powers of x cancel, since x⁻⁴ · x³ · x = 1, which leaves (y² + x)⁴ = C|(y² + 2x)³|. That is option (a).
The idea to keep. When y shows up only as y² next to a y dy, substitute t = y² first. A hidden homogeneous equation often appears, and then t = vx does the rest.
Answer from the NTA final answer key. Solution by PrepWiser.