Test: Differential Equations: Reducible Separable Method(21 Oct) - JEE MCQ

Order = 3 and Degree = 1
The order of a differential equation is determined by the highest-order derivative; the degree is determined by the highest power on a variable. The higher the order of the differential equation, the more arbitrary constants need to be added to the general solution.

The order of a differential equation representing a family of curves is same as the number of arbitrary constants present in the equation corresponding to the family of curves.

x³dx + (y + 1)dy = 0
=> d(x⁴/4) + d((y + 1)³/3) = 0
=> d(x⁴/4 + ((y + 1)³/3)) = 0
=> (x⁴/4 + ((y + 1)³/3)) = constt
=> 12 *(x⁴/4 + ((y + 1)³/3)) = 12 * constt
=> 3x⁴ + 4(y + 1)³ = constt

(x−y)dy/dx = x + 2y
⇒ dy/dx = (x + 2y)/(x−y)
F (x,y) = (x + 2y)/(x−y)
F(Ax, Ay) = (Ax + 2Ay)/(Ax−Ay) 
= A(x + 2y)/A(x−y) 
= (x + 2y)/(x − y) = F (x,y)
Hence, the equation is homogenous.

xdy/dx = -coty
dy/dx = - coty/x
dy/coty = - dx/x
tany dy = - dx/x 
∫(siny/cosy)dy = - ∫dx/x…………..(1)
Put t = cosy
dt = -sinydy
Put the value of dt in eq(1)
-∫dt/t = -∫dx/x
log |t| = log |x| + log c
log |cosy| = log(c.|x|)
|cos y| = c.|x|....................(2)
As y = π/4, x = (2)^½
cos π/4 = c*(2)^½
1/(2)^½ = c*(2)^½ 
c = ½
Put the value of ‘c’ in eq(2)
cos y = x/2    => x = 2cosy

dy/dx = e^ax cos by 
∫dy/cos by = ∫e^ax dx
∫sec by dy = ∫e^ax dx
= (log| sec by + tan by|)/b = e^ax /a + c
= a(log| sec by + tan by|) = be^ax + c

dy/dx = x logx
=> ∫dy = ∫x logx dx
y = logx . x²/2 - ∫1/x . x²/2 dx + c
y = x²/2 log x -½ ∫x dx + c
y = x²/2 log x - ½ . x²/2 + c
y = x²/2 log x - x²/4 + c

 d²y/dx² = xe^x
d²y/dx² =  ∫xe^x
d/dx(dy/dx) = [xe^x]
Integrating, we get
dy/dx = (x-1)e^x+ c1
Again integrating, y = (x-2)e^x + c1x + c2

 e^dy/dx = x + 1
⇒ dy/dx = log(x+1)
⇒ dy = log(x+1)dx
Integrating both sides, we get
⇒ ∫dy =  ∫log(x+1)dx
y = log(x + 1)  ∫1dx - ∫[d/dx{log(x + 1)} ∫1dx]dx
y = xlog(x + 1) -  ∫1/(x+1)xdx
y = xlog(x + 1) -  ∫(1 - 1/(x + 1)dx
y = xlog(x + 1) -  ∫dx +  ∫1/(x+1)dx
y = xlog(x + 1) - x + log|x + 1| + c
y = (x + 1)log|x + 1| - x + c
Here c is 3
y = (x + 1)log|x + 1| - x + 3