Homogeneous Equations Questions and Answers

This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses 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) \(\frac{ln⁡x}{x} – \frac{ln⁡y}{y} = c\)
View Answer

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) \( \frac{e^{2y}}{2} = \frac{e^{3x}}{3} + \frac{x^3}{3} + c\)
View Answer

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
View Answer

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) \(\frac{1}{2\sqrt{2}} tan^{-1}⁡(\frac{4x+2y+1}{\sqrt{2}})=x+c \)
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
View Answer

Answer: a
Explanation: \(\frac{dy}{dx} = (4x+2y+1)^2\)
here we use substitution for \( 4x+2y+1 = t→4 + 2\frac{dy}{dx} = \frac{dt}{dx} \rightarrow \frac{dy}{dx} = \frac{1}{2} \frac{dt}{dx} – 2\)
\(\frac{1}{2} \frac{dt}{dx} – 2 = t^2\)
\(\frac{dt}{dx}=2t^2+4\)
separating the variable and integrating
\(\int \frac{1}{2t^2+4} \,dt = \int dx\)
\(\frac{1}{2\sqrt{2}} tan^{-1}⁡ ⁡\frac{t}{\sqrt{2}} = x + c\)
\(\frac{1}{2\sqrt{2}} tan^{-1}⁡(\frac{4x+2y+1}{\sqrt{2}})=x+c \) 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
View Answer

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
\(\frac{y}{1+y} \,dy = \frac{1+x}{x} \,dx\)
\(\int\frac{y}{1+y} \,dy = \int \frac{1+x}{x} \,x\)
\(\int\frac{(1+y)-1}{1+y} \,dy = \int\frac{1}{x} \,dx + \int1 \,dx\)
\(\int1 \,dy – \int\frac{1}{1+y} \,dy – log \,x – x = c\)
y – log(1+y) – log x – x = c
(y-x) – log(x(1+y)) = c is the solution.

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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
View Answer

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 \(y = vx\rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} = \frac{1+v^2}{3v}\)
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) \(sin(\frac{y}{x}) = xc\)
View Answer

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
View Answer

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
\(x\frac{dv}{dx} = \frac{v^2-2v-1}{v^2+2v-1} – v = -\frac{(v^3+v^2+v+1)}{v^2+2v-1} = -\frac{(v^2+1)(v+1)}{v^2+2v-1}\)
\(\int \frac{v^2+2v-1}{(v^2+1)(v+1)} \,dv = \int \frac{-1}{x} \,dx\)
\(\int\frac{2v(v+1)-(v^2+1)}{(v^2+1)(v+1)} \,dv = log \,c – log \,x\)
\(\int(\frac{2v}{v^2+1}-\frac{1}{v+1}) \,dv = log \,c – log \,x\)
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.

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