Test: Homogeneous Differential Equations - JEE MCQ

 dy/dx = y/x - (y²/x² + 1)^½…………………(1)
the equation is homogenous,so y = Vx
dy/dx = V + xdv/dx 
Put the value of dy/dx in eq(1)
V + xdv/dx = V - (V² + 1)^½
= ∫-dv(v² + 1)^½ = ∫dx/x
= log|v + (v² + 1)^½| = -logx + logc
= log|y/x + ((y/x)² + 1)^½| =  log c/x 
=|y/x + ((y/x)² + 1)^½| =  c/x

dy/dx = (x² + 3y²)/2xy………….(1)
Let y = vx
dy/dx = v + xdv/dx
Substitute the value of y and dy/dx in (1)
v + x dv/dx = (1+3v²)/2v
x dv/dx = (1+3v²)/2v - v
x dv/dx = (1 + 3v² - 2v²)/2
x dv/dx = (1+ v²)/2v
2v/(1+v²) dv = dx/x…………(2)
Integrating both the sides
∫2v/(1+v²) dv = ∫dx/x
Put t = 1 + v²
dt = 2vdv
∫dt/t  = ∫dx/x
=> log|t| = log|x| + log|c|
=> log|t/x| = log|c|
t/x = +- c
(1+v²)/x = +-c
(1 + (y²)/(x²))/x = +-c
x² + y² = Cx³……….(3)
y(1) = 3
1 + 9 = c(1)³
c = 10
From eq(3), we get x²+ y² = 10x³

Given x dy/dx = y(log y - log x+1)
logy-logx=log(y/x)
dy/dx=y/x(log(y/x) +1)
substituting y=vx
dy/dx=v+xdv/dx
=>dv/(vlogv)=dx/x, integrating we get
log(logv)=log(cx)
log(y/x) = cx
=> y/x = e^(cx)
=> y = xe^(cx)

This is clearly a Homogenous differential equation, as RHS is expressed only in terms of y/x.
To Solve this, Lety/x=t
⟹ y=xt
⟹ dy/dx = t+xdt/dx
By (i)
⟹ tant+t = t+xdt/dx
⟹ tant=xdt/dx
⟹ dx/x=dt/tant
Integrating both sides,
⟹ ∫dx/x = ∫cot tdt
⟹logx = log(sint)+logC
⟹ logx=log(C siny/x)
⟹ x = Csin(y/x)

 x -(xy)^½ dy = ydx
[x-(xy)^½] - y = dx/dy
dx/dy = x/y - [(xy)^½]/y……………….(1)
Let V = x/y
x = Vy
dx/dy = V + ydv/dy……..(2)
V + ydv/dy = V - (V)^½
ydv/dy = -(V)^½
dv/(V)^½ = -dy/y
Integrating both the sides, we get
(V^(-½+1))/(-½ + 1) = -log y + c
2(V)^½ = -log y + c
2(y/x)^½ = -log y + c
2(x)^½ = (y)^½(-log y + c)
2(x)^½ = (y)¹/2log y + c(y)^½ 
(x)^½ = [(y)^½]/2 log y + [c(y)^½]/2

y/x cos y/x dx − (x/y sin y/x + cosy/x) dy = 0
⇒dy/dx = (y/x cos y/x)/(x/y sin y/x + cos y/x)
This is a homogeneous differential equation.
Putting y = vx and dy/dx = v+xdv/dx, we have v+xdv/dx = (v²cosv)/(sinv+vcosv)
⇒ xdv/dx = (v²cosv)/(sinv+vcosv) - v
⇒x dv/dx = (v² cosv - v² cosv - vsinv )/(sinv+vcosv) 
⇒ x dv/dx = - [v sinv/(sinv + vcosv)]
⇒ ∫[(sinv + vcosv)/v sinv]dv = ∫dx/x
Integrating both the sides, we get
∫(cot v + 1/v)dv = - ln(x) + c
ln(sin v) + ln(v) = -ln(x) + c
ln(sin(y/x) + ln(y/x) + ln(x) = c
ln(y/x sin y/x * x) + c
y sin(y/x) = c

1. Let y = mx + c be the equation of all the straight lines touching the circle.

   Given : The equation of the circle is x² + y² = r²----------> (1)

   The tangent to the circle is c² = r²(1+m²)

   c = r√(1+m²)

   we know that y = mx + c---------->(2)

   y = mx + r√(1+m²) ---------->(3)

   y - mx = r√(1+m²)

   Differentiating wrt x we get dy/dx -m =0

   dy/dx = m

   Substituting this in equation (3)

   y - (dy/dx . x) = r√(1+(dy/dx)²)

   Squaring on both sides, we get

   y² - (dy/dx . x)² = [ r√(1+(dy/dx)²)]²

   [y - x(dy/dx)]² = r² (1+(dy/dx))² is the required differential equation.

   Answer: The differential equation of all the straight lines touching the circle x² + y² = r² is [y - x(dy/dx)]² = r² (1+(dy/dx))²​​​​​​

x²dy + (xy + y²) dx = 0
⇒ x² dy = - (xy + y²)dx
⇒ .......(i)
Let y = vx
Differentiating w.r.t x we get

Substituting the value of y and dy/dx in equation (1), we get:

Integrating both sides, we get:


......(ii)
Now, it is given that y = 1 at x = 1.

Substituting D = 1/3 in equation (2), we get

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