The solution of the differential equation dy dx=-( x^2+3y^2 3x^2+y^2),y(1)=0 is

Given,

Differential equation,

\frac{dy}{dx}=-(\frac{{x}^{2}+3{y}^{2}}{3{x}^{2}+{y}^{2}})&y(1)=0

Now let $y=vx$y=vx

$$v+x\frac{dv}{dx}=\frac{dy}{dx}$$v+xdxdv​=dxdy​

$$\Rightarrow v+x\frac{dv}{dx}=-(\frac{1+3{v}^{2}}{3+{v}^{2}})$$⇒v+xdxdv​=−(3+v21+3v2​)

$$\Rightarrow x\frac{dv}{dx}=-\frac{{(v+1)}^{3}}{3+{v}^{2}}$$⇒xdxdv​=−3+v2(v+1)3​

$$\Rightarrow \frac{(3+{v}^{2})dv}{{(v+1)}^{3}}+\frac{dx}{x}=0$$⇒(v+1)3(3+v2)dv​+xdx​=0

$$\Rightarrow \int \frac{4dv}{{(v+1)}^{3}}+\int \frac{dv}{v+1}-\int \frac{2dv}{{(v+1)}^{2}}+\int \frac{dx}{x}=0$$⇒∫(v+1)34dv​+∫v+1dv​−∫(v+1)22dv​+∫xdx​=0

$$\Rightarrow \frac{-2}{{(v+1)}^{2}}+\mathrm{ln}|v+1|+\frac{2}{v+1}+\mathrm{ln}|x|=c$$⇒(v+1)2−2​+ln∣v+1∣+v+12​+ln∣x∣=c

$$\Rightarrow \frac{-2{x}^{2}}{{(x+y)}^{2}}+\mathrm{ln}|\frac{x+y}{x}|+\frac{2x}{x+y}+\mathrm{ln}|x|=c$$⇒(x+y)2−2x2​+ln∣xx+y​∣+x+y2x​+ln∣x∣=c

$$\Rightarrow \frac{2xy}{{(x+y)}^{2}}+\mathrm{ln}|x+y|=c$$⇒(x+y)22xy​+ln∣x+y∣=c

$$\because y(1)=0$$∵y(1)=0

$\therefore c=0$∴c=0, as $x=1,y=0$x=1,y=0

$$\Rightarrow \frac{2xy}{{(x+y)}^{2}}+\mathrm{ln}|x+y|=0$$⇒(x+y)22xy​+ln∣x+y∣=0