This set of 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
View Answer
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 \(x-2y = t \rightarrow 1 – 2\frac{dy}{dx} = \frac{dt}{dx}\)
\(1- \frac{dt}{dx} = 2\left(\frac{3t+7}{t+4}\right) \rightarrow \frac{dt}{dx} = \frac{t+4-6t-14}{t+4} =\frac{-5(t+2)}{t+4}\)
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
View Answer
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
\(V + X\frac{dV}{dX} = \frac{7-3V}{-3+7V}\)……separating the variables we get & integrating
\(X\frac{dV}{dX} = \frac{7-3V}{-3+7V} – V = \frac{7-7V^2}{7V-3}\)
\( \int \frac{7V-3}{1-V^2} \,dV = 7\int \frac{1}{X} \,dX\)
\( \int \frac{7V}{1-V^2} \,dV – \int \frac{3}{1-V^2} \,dV\) = 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) \((x^2+y^2)^1 = c tan\left(\frac{y+0.5}{x+0.5}\right)\)
View Answer
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-1v-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
View Answer
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
\(v + X\frac{dv}{dX} = \frac{1+2v}{2+v} \rightarrow X \frac{dv}{dX} = \frac{1-v^2}{2+v}\)
or
\( \int(\frac{2+v}{1-v^2}) dv = \int \frac{1}{X} \,dX \rightarrow \int(\frac{2}{1-v^2} + \frac{v}{1-v^2}) \,dv = log X + log c\)
\(log\left(\frac{1+v}{1-v}\right) – 0.5 log (1-v^2) = log X + log c\)
\(log \left(\frac{X+Y}{X-Y}\right) – 0.5 log (X^2-Y^2) + log X = log X + log c\)
\(log \left(\frac{X+Y}{X-Y}.\frac{1}{\sqrt{X^2-Y^2}}\right) = log c\)
\(\sqrt{X+Y}\) = (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) \(π-tan{-1} \frac{y+3}{x+2} – log\sqrt{(x+2)^2+(y+3)^2}=c\)
b) \(-tan{-1} \frac{y+3}{x+2} – log\sqrt{(x+2)^2+(y+3)^2}=c\)
c) \(tan{-1} \frac{y+3}{x+2} – log\sqrt{x+y+5}=c\)
d) \(-cot{-1} \frac{y+3}{x+2} – log\sqrt{x+y+5}=c\)
View Answer
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
\(V + X\frac{dV}{dX} = \frac{v-1}{v+1} \rightarrow X\frac{dV}{dX} = \frac{-(1+v^2)}{1+v}\)
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-1v-0.5 log(1+v2)=log x +c
\(-tan^{-1} \frac{Y}{X} – log\sqrt{X^2+Y^2} = c\)
\(-tan{-1} \frac{y+3}{x+2} – log\sqrt{(x+2)^2+(y+3)^2}=c\) is the solution.
Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.
To practice all areas of Ordinary Differential Equations for Interviews, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
- Practice Probability and Statistics MCQ
- Check Engineering Mathematics Books
- Apply for 1st Year Engineering Internship
- Practice Numerical Methods MCQ
