250+ TOP MCQs on Separable and Homogeneous Equations and Answers

Ordinary Differential Equations Multiple Choice Questions on “Separable and Homogeneous Equations”.

1. Solution of the differential equation (frac{dy}{dx} = frac{y(x-y ln⁡y)}{x(x ln x-y)}) is _____________
a) (frac{x ln⁡x+y ln⁡y}{xy} = c)
b) (frac{x ln⁡x-y ln⁡y}{xy} = c)
c) (frac{ln⁡x}{x} + frac{ln⁡y}{y} = c)
d) (frac{ln⁡x}{x} – frac{ln⁡y}{y} = c)
Answer: a
Explanation: (frac{dy}{dx} = frac{y(x-y ln⁡y)}{x(x ln x-y)})
–> x² ln⁡x dy-xy dy=xy dx – y² ln⁡y dx …….dividing by x² y² then
(frac{ln⁡y}{y^2} ,dy, – frac{1}{xy} ,dy, = frac{1}{xy} ,dx, – frac{ln⁡y}{x^2} ,dx)
((ln x(frac{1}{-y^2}dy) + frac{1}{xy} ,dx) + (ln y(frac{1}{-x^2}dx) + frac{1}{xy} ,dy) = 0 )
(d(frac{ln⁡x}{y}) + d(frac{ln⁡y}{x}) = 0)
on integrating we get
(int d(frac{ln⁡x}{y}) + int d(frac{ln⁡y}{x}))
( frac{ln⁡x}{y} + frac{ln⁡y}{x}) = c…. where c is a constant of integration.

2. Solution of the differential equation (frac{dy}{dx} = e^{3x-2y} + x^2 e^{-2y}) is ______
a) ( frac{e^{2y}}{3} = frac{e^{3x}}{3} + frac{x^2}{2} + c)
b) ( frac{e^{3y} (e^{2x}+x^3)}{6} + c)
c) (frac{e^{2y} (e^{3x}+x^3)}{6} + c)
d) ( frac{e^{2y}}{2} = frac{e^{3x}}{3} + frac{x^3}{3} + c)
Answer: d
Explanation: (frac{dy}{dx} = e^{3x-2y} + x^2 e^{-2y})
(frac{dy}{dx} = e^{-2y} (e^{3x} + x^2))
separating the variable
e^2y dy = (e^3x+x²)dx…..integrating
∫ e^2y dy =∫ (e^3x+x²)dx
( frac{e^{2y}}{2} = frac{e^{3x}}{3} + frac{x^3}{3} + c).

3. Solution of the differential equation sec² x tan⁡y dx + sec² y tan⁡x dy=0 is _______
a) (sec x. sec y)=k
b) (sec x .tany)=k
c) (tan x. tany)=k
d) (sec x .tan x)+(sec y .tan y)=k
Answer: c
Explanation: sec² x tan⁡y dx + sec² y tan⁡x dy=0
dividing throughout by tan y.tan x we get
(frac{sec^2 x}{tan⁡x} ,dx + frac{sec^2 y}{tan⁡y} ,dy=0)……separating the variable
now integrating we get (int frac{sec^2 x}{tan⁡x} ,dx + int frac{sec^2 y}{tan⁡y} ,dy=c)
substituting tan x = t & tan y=p→sec² x dx=dt & sec² y dy=dp
( rightarrow int frac{1}{t} ,dt + int frac{1}{p} ,dp=c)
log t + log p = c –>log(tan x)+log(tan y) = c = log k….since it is an unknown constant
log(tan x .tan y) = log k
(tan x tan y) = k is the solution.

4. Solution of the differential equation (frac{dy}{dx} = (4x+2y+1)^2) is ______
a) (frac{1}{2sqrt{2}} tan^{-1}⁡(frac{4x+2y+1}{sqrt{2}})=x+c )
b) (frac{1}{sqrt{2}} cot^{-1}⁡(4x+2y+1)=x+c )
c) (frac{1}{sqrt{2}} tan^{-1}⁡⁡(frac{4x+2y+1}{sqrt{2}})=c )
d) cot^-1⁡⁡⁡(4x+2y+1)=x+c
Answer: a
Explanation: (frac{dy}{dx} = (4x+2y+1)^2)
here we use substitution for ( 4x+2y+1 = t→4 + 2frac{dy}{dx} = frac{dt}{dx} rightarrow frac{dy}{dx} = frac{1}{2} frac{dt}{dx} – 2)
(frac{1}{2} frac{dt}{dx} – 2 = t^2)
(frac{dt}{dx}=2t^2+4)
separating the variable and integrating
(int frac{1}{2t^2+4} ,dt = int dx)
(frac{1}{2sqrt{2}} tan^{-1}⁡ ⁡frac{t}{sqrt{2}} = x + c)
(frac{1}{2sqrt{2}} tan^{-1}⁡(frac{4x+2y+1}{sqrt{2}})=x+c ) is the solution.

5. Solution of the differential equation (xy frac{dy}{dx} = 1+x+y+xy) is ______
a) (y-x)-log(x(1+y))=c
b) log(x(1+y))=c
c) (y+x)-log(x)=c
d) (y-x)-log(y(1+x))=c
Answer: a
Explanation: (xy frac{dy}{dx} = 1+x+y+xy)
(xy frac{dy}{dx} = (1+x)+y(1+x)=(1+x)(1+y))
separating the variables & hence integrating
(frac{y}{1+y} ,dy = frac{1+x}{x} ,dx)
(intfrac{y}{1+y} ,dy = int frac{1+x}{x} ,x)
(intfrac{(1+y)-1}{1+y} ,dy = intfrac{1}{x} ,dx + int1 ,dx)
(int1 ,dy – intfrac{1}{1+y} ,dy – log ,x – x = c)
y – log(1+y) – log x – x = c
(y-x) – log(x(1+y)) = c is the solution.

6. Solve the differential equation (frac{dy}{dx} = frac{x^2+y^2}{3xy}) is _______
a) xp=(x²+2y²)^-3
b) x² p=(x²-2y²)³
c) x⁴ p=(x²-2y²)^-3
d) x⁶ p=(x²+2y²)³
Answer: b
Explanation: (frac{dy}{dx} = frac{x^2+y^2}{3xy} = frac{1+frac{y^2}{x^2}}{3 frac{y}{x}}) we can clearly see that it is an homogeneous equation
hence substituting (y = vxrightarrow frac{dy}{dx} = v + x frac{dv}{dx} = frac{1+v^2}{3v})
separating the variables and integrating
(x frac{dv}{dx} = frac{1+v^2}{3v} – v = frac{1-2v^2}{3v})
(int frac{3v}{1-2v^2} ,dv = int frac{1}{x} dx)…….substituting 1-2v²=t→-4v dv=dt we get
(frac{-3}{4} log⁡ ,t = log⁡ ,x + log⁡ ,c rightarrow frac{-3}{4} log⁡(1-2v^2) = log ⁡,cx…..but ,v = frac{y}{x})
(-3log⁡(frac{x^2-2y^2}{x^2})=4log⁡ ,cx rightarrow log⁡(frac{x^2-2y^2}{x^2})^{-3} = log⁡ ,kx^4 )
(frac{x^6}{(x^2-2y^2)^3} = kx^4 rightarrow x^2 ,p = (x^2-2y^2)^3) is the solution where p is constant.

7. The solution of differential equation (frac{dy}{dx} = frac{y}{x} + tan⁡frac{y}{x}) is ______
a) (cot(frac{y}{x}) = xc)
b) (cos(frac{y}{x}) = xc)
c) (sec^2(frac{y}{x}) = xc)
d) (sin(frac{y}{x}) = xc)
Answer: d
Explanation: (frac{dy}{dx} = frac{y}{x} + tan⁡frac{y}{x}) we can clearly see that it is an homogeneous equation
substituting (y = vx rightarrow frac{dy}{dx} = v + x frac{dv}{dx} = v + tan ,v)
separating the variables and integrating we get
(int frac{1}{tan⁡v} ,dv = int frac{1}{x} dx)
log(sin v) = log x + log c
(sin ,v = xc rightarrow sin(frac{y}{x}) = xc) is the solution where c is constant.

8. Particular solution of the differential equation (frac{dy}{dx} = frac{y^2-2xy-x^2}{y^2+2xy-x^2}) given y=-1 at x=1.
a) y=x
b) y+x=2
c) y=-x
d) y-x=2
Answer: c
Explanation: (frac{dy}{dx} = frac{y^2-2xy-x^2}{y^2+2xy-x^2} = frac{frac{y^2}{x^2} – frac{2y}{x} – 1}{frac{y^2}{x^2} + frac{2y}{x} -1} )……. is a homogeneous equation
thus put (y = vx rightarrow frac{dy}{dx} = v + x frac{dv}{dx} = frac{v^2-2v-1}{v^2+2v-1})
separating the variables and integrating we get
(xfrac{dv}{dx} = frac{v^2-2v-1}{v^2+2v-1} – v = -frac{(v^3+v^2+v+1)}{v^2+2v-1} = -frac{(v^2+1)(v+1)}{v^2+2v-1})
(int frac{v^2+2v-1}{(v^2+1)(v+1)} ,dv = int frac{-1}{x} ,dx)
(intfrac{2v(v+1)-(v^2+1)}{(v^2+1)(v+1)} ,dv = log ,c – log ,x)
(int(frac{2v}{v^2+1}-frac{1}{v+1}) ,dv = log ,c – log ,x)
log(v²+1) – log(v+1) + log x = log c –> (frac{(v^2+1)x}{(v+1)}=c)
(rightarrow frac{x^2+y^2}{x+y} = c rightarrow k(x^2+y^2)=(x+y)) where k=1/c
at x=1, y=-1 substituting we get 2k=0→k=0
thus the particular solution is y=-x.

Global Education & Learning Series – Ordinary Differential Equations.

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