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  • The solution of the differential equation <img alt="\frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right), y(1)=0" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bd%20y%7

The solution of the differential equation \frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right), y(1)=0 is

Option: 1

\log _{e}|x+y|-\frac{x y}{(x+y)^{2}}=0


Option: 2

\log _{e}|x+y|+\frac{2 x y}{(x+y)^{2}}=0


Option: 3

\log _{e}|x+y|-\frac{2 x y}{(x+y)^{2}}=0


Option: 4

\log _{e}|x+y|+\frac{x y}{(x+y)^{2}}=0


Answers (1)

best_answer

y=v x
\frac{d y}{d x}=v+\frac{x d v}{d x}
v+\frac{x d v}{d x}=-\frac{x^{2}+3 v^{2} x^{2}}{3 x^{2}+v^{2} x^{2}}
x \frac{d v}{d x}=-\frac{1+3 v^{2}}{3+v^{2}}-v
x \frac{d v}{d x}=-\frac{1+3 v^{2}+3 v+v^{3}}{3+v^{2}}

\int \frac{3+v^{2}}{1+3 v^{2}+3 v+v^{3}} d v=-\int \frac{d x}{x}
\Rightarrow \int \frac{3+v^{2}}{(1+v)^{3}} d v=-\ln x+C

\text { Let } \quad v+1=t
              dv= dt
\int \frac{3+(t-1)^{2}}{t^{3}} d t=-\ln x+C
\Rightarrow \int \frac{t^{2}-2 t+4}{t^{3}} d t
\Rightarrow \int\left(\frac{1}{t}-\frac{2}{t^{2}}+\frac{4}{t^{3}}\right) d t=-\ln x+C
\Rightarrow \ln t+\frac{2}{t}-\frac{4}{2 t^{2}}=-\ln x+C
\Rightarrow \ln \left(\frac{y}{x}+1\right)^{-1} \frac{2}{\frac{y}{x}+1}-\frac{4}{2(y / x+1)^{2}}=-\ln x+C
\Rightarrow \ln \left(\frac{y+x}{x}\right)+\frac{2 x}{y+x}-\frac{2 x^{2}}{(x+y)^{2}}=-\ln x+C
\Rightarrow \ln \left(\frac{y+x}{x}\right)+\frac{2 x}{y+x}-\frac{2 x^{2}}{(x+y)^{2}}=-\ln x+C
\Rightarrow \ln |x+y|+\frac{2 x}{(x+y)^{2}}(x+y-x)=C
\Rightarrow \ln |x+y|+\frac{2 x y}{(x+y)^{2}}=C
 

Posted by

Rishabh

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