Question

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

This question was posed to me during an online exam.

I want to ask this question from Reducible to Homogenous Form in section Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct option is (d) p=(y+x-1)^5 (y-x+1)^2

Easy 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-V^2=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 X^7=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.