250+ TOP MCQs on Reducible to Homogenous Form and Answers

Ordinary Differential Equations Interview Questions and Answers focuses on “Reducible to Homogenous Form”.

1. Solution of the differential equation (frac{dy}{dx} = frac{3x-6y+7}{x-2y+4}) is _____
a) log(x-2y+4)²=c
b) 6x-2y+log(x-2y+2)²=c
c) x-2y+log(x-2y+6)²=c
d) -5x+log(x-2y+3)²=c
Answer: b
Explanation: (frac{dy}{dx} = frac{3x-6y+7}{x-2y+4} rightarrow frac{dy}{dx} = frac{3(x-2y)+7}{(x-2y)+4})…..here coefficient of x and y in the numerator & denominator are proportional hence substituting (x-2y = t rightarrow 1 – 2frac{dy}{dx} = frac{dt}{dx})
(1- frac{dt}{dx} = 2left(frac{3t+7}{t+4}right) rightarrow frac{dt}{dx} = frac{t+4-6t-14}{t+4} =frac{-5(t+2)}{t+4})
separating the variables and hence integrating
( int frac{t+4}{t+2} ,dt = int -5 ,dx rightarrow int left(1 + frac{2}{t+2}right) ,dt = -5x + c)
t + 2 log(t+2) = -5x + c –> x – 2y + 2 log(x-2y+2) + 5x = c
6x-2y+log(x-2y+2)² = c is the solution.

2. Solution of the differential equation (3y-7x+7)dx+(7y-3x+3)dy=0 is ______
a) p=(y+x)⁵ (y-x)²
b) p=(y+x+2)⁵ (y-x+2)²
c) p=(y+x)² (y-x)⁵
d) p=(y+x-1)⁵ (y-x+1)²
Answer: d
Explanation: (3y-7x+7)dx + (7y-3x+3)dy = 0 –> (frac{dy}{dx} = frac{7x-3y-7}{-3x+7y+3})
substituting x=X+h, y=Y+k where (h,k) will satisfy the equation
– 3y+7x-7=0 & 7y-3x+3=0…..(1)
after substitution we get (frac{dY}{dX} = frac{(7X-3Y)+(7h- 3k-7)}{(-3X+7Y)+(-3h+7k+3)})…..(2)
from (1) we can write 7h- 3k-7=0 & -3h+7k+3=0 solving for
h & k we get h=1 & k=0 (2) can be written as (frac{dY}{dX} = frac{7X-3Y}{-3X+7Y})
(frac{dY}{dX} = frac{7X-3Y}{-3X+7Y})…..it is a homogenous equation hence substituting Y=VX
(V + Xfrac{dV}{dX} = frac{7-3V}{-3+7V})……separating the variables we get & integrating
(Xfrac{dV}{dX} = frac{7-3V}{-3+7V} – V = frac{7-7V^2}{7V-3})
( int frac{7V-3}{1-V^2} ,dV = 7int frac{1}{X} ,dX)
( int frac{7V}{1-V^2} ,dV – int frac{3}{1-V^2} ,dV) = 7log X + 7log c……7log c=log k
substituting 1-V²=t –> -2V dV=dt
( int frac{7}{-2t} ,dt + frac{3}{2} log(frac{V-1}{V+1})- 7 log X = log k)
(frac{7}{-2t} log(1-V^2) + frac{3}{2} log(frac{V-1}{V+1}) – 7 log X = log k)
(log(1-V^2)^{frac{7}{-2t}} + log(frac{V-1}{V+1})^{frac{3}{2}} + log(X^{-7}) = log k)
((1-V^2)^{frac{7}{-2t}} (frac{V-1}{V+1})^{frac{3}{2}} X^{-7} = k)
(1+V)⁵ (V-1)² X⁷=p where 1/k = p
p=(y+x-1)⁵ (y-x+1)² is the solution…after substituting back V=Y/X & Y=y, X=x-1.

3. Solution of the differential equation (x-y)dy=(x+y+1)dx is _____
a) ( sqrt{x^2+y^2} = e^{c tan^{-1}⁡left(frac{y+0.5}{x+0.5}right)})
b) ((x^2+y^2)^1.5 = c cot^{-1} ⁡left(frac{y+0.5}{x+0.5}right))
c) ((x^2+y^2)^2 = e^{c cot⁡⁡left(frac{y+0.5}{x+0.5}right)})
d) ((x^2+y^2)^1 = c tan⁡left(frac{y+0.5}{x+0.5}right))
Answer: a
Explanation: (frac{dy}{dx} = frac{x+y+1}{x-y}), by substituting x=X+h, y=Y+k
w.k.t (h,k) satisfies the equations as h+k+1=0 & h-k=0 –> h=k=-0.5
and hence the differential equation changes to the form (frac{dy}{dx}=frac{X+Y}{X-Y} ) is a homogenous equation thus put (v = frac{Y}{X} rightarrow ,v + X frac{dv}{dX} = frac{1+v}{1-v})
(X frac{dv}{dX} = frac{1+v^2}{1-v} rightarrow int frac{1-v}{1+v^2} ,dv = ∫frac{1}{X} ,dx = int(frac{1}{1+v^2} – frac{v}{1+v^2}) ,dv = log X + c)
tan^-1⁡v-0.5log (1+v²) = log X+c –> ( tan^{-1} frac{⁡Y}{X} – log⁡(sqrt{X^2+Y^2}) = c)
(sqrt{X^2+Y^2} = e^{c tan^{-1}⁡left(frac{y+0.5}{x+0.5}right)})……from the equation X=x+0.5 & Y=y+0.5.

4. Solution of the differential equation (frac{dy}{dx} = frac{x+2y-3}{2x+y-3}) is _____
a) (x+y)=c(x-y)²
b) (x+y-2)=c(x-y)³
c) (2x+y)=c(x+2y)²
d) (2x+y-3)=c(x-2y-3)³
Answer: b
Explanation: (frac{dy}{dx} = frac{x+2y-3}{2x+y-3}) –>x=X+h, y=Y+k
h+2k-3=0 & 2h+k-3=0 solving we get h=1=k
new differential equation is (frac{dY}{dX} = frac{X+2Y}{2X+Y}), put Y=vX
(v + Xfrac{dv}{dX} = frac{1+2v}{2+v} rightarrow X frac{dv}{dX} = frac{1-v^2}{2+v})
or
( int(frac{2+v}{1-v^2}) dv = int frac{1}{X} ,dX rightarrow int(frac{2}{1-v^2} + frac{v}{1-v^2}) ,dv = log X + log c)
(logleft(frac{1+v}{1-v}right) – 0.5 log (1-v^2) = log X + log c)
(log left(frac{X+Y}{X-Y}right) – 0.5 log (X^2-Y^2) + log X = log X + log c)
(log left(frac{X+Y}{X-Y}.frac{1}{sqrt{X^2-Y^2}}right) = log c)
(sqrt{X+Y}) = (X-Y)^1.5 c or (X+Y)=(X-Y)^3c
(x+y-2)=c(x-y)³ is the solution.

5. Solution of the differential equation (frac{dy}{dx} = frac{y-x+1}{y+x+5}) is ______
a) (π-tan{-1}⁡ frac{y+3}{x+2} – logsqrt{(x+2)^2+(y+3)^2}=c)
b) (-tan{-1}⁡⁡ frac{y+3}{x+2} – logsqrt{(x+2)^2+(y+3)^2}=c)
c) (tan{-1}⁡⁡ frac{y+3}{x+2} – logsqrt{x+y+5}=c)
d) (-cot{-1}⁡⁡ frac{y+3}{x+2} – logsqrt{x+y+5}=c)
Answer: b
Explanation: (frac{dy}{dx} = frac{y-x+1}{y+x+5}) –> x=X+h, y=Y+k
h+k+5=0 & k-h+1=0 solving we get h=-2, k=-3
(frac{dY}{dX} = frac{Y-X}{Y+X}), put Y=vX we get an new equation
(V + Xfrac{dV}{dX} = frac{v-1}{v+1} rightarrow Xfrac{dV}{dX} = frac{-(1+v^2)}{1+v})
or
(-int frac{1+v}{1+v^2} dv = int frac{1}{X} dX = int frac{1}{1+v^2} + frac{v}{1+v^2} ,dv)
-tan^-1⁡v-0.5 log(1+v²)=log x +c
(-tan^{-1} frac{⁡Y}{X} – logsqrt{X^2+Y^2} = c)
(-tan{-1}⁡⁡ frac{y+3}{x+2} – logsqrt{(x+2)^2+(y+3)^2}=c) is the solution.

Global Education & Learning Series – Ordinary Differential Equations.

To practice all areas of Ordinary Differential Equations for Interviews,