Differential Equations Question 8
Question 8 - 30 January - Shift 2
The solution of the differential equation
$\frac{d y}{d x}=-(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}), y(1)=0$ is
(1) $\log _{e}|x+y|-\frac{x y}{(x+y)^{2}}=0$
(2) $\log _{e}|x+y|+\frac{x y}{(x+y)^{2}}=0$
(3) $\log _{e}|x+y|+\frac{2 x y}{(x+y)^{2}}=0$
(4) $\log _{e}|x+y|-\frac{2 x y}{(x+y)^{2}}=0$
Show Answer
Answer: (3)
Solution:
Formula: Homogeneous equations
Put $y=v x$
$v+x \frac{d v}{d x}=-(\frac{1+3 v^{2}}{3+v^{2}})$
$x \frac{d v}{d x}=-\frac{(v+1)^{3}}{3+v^{2}}$
$\frac{(3+v^{2}) d v}{(v+1)^{3}}+\frac{d x}{x}=0$
$\int \frac{4 d v}{(v+1)^{3}}+\int \frac{d v}{v+1}-\int \frac{2 d v}{(v+1)^{2}}+\int \frac{d x}{x}=0$
$\frac{-2}{(v+1)^{2}}+\ln (v+1)+\frac{2}{v+1}+\ln x=c$
$\frac{-2 x^{2}}{(x+y)^{2}}+\ln (\frac{x+y}{x})+\frac{2 x}{x+y}+\ln x=c$
$\frac{2 xy}{(x+y)^{2}}+\ln (x+y)=c$
$\therefore c=0$, as $x=1, y=0$
$\frac{2 x y}{x+y)^{2}}+\ln (x+y)=0$
