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  • Solve for particular solution. x^2 dy + (xy + y^2) dy = 0; y = 1 when x = 1

Solve for particular solution.

    Q12. x^2dy + (xy + y^2)dx = 0; y =1\ \textup{when}\ x = 1

Answers (1)

best_answer

\frac{dy}{dx}= \frac{-(xy+y^{2})}{x^{2}} = F(x,y)...............................(i)

F(\mu x, \mu y)=\frac{-\mu^{2}(xy+(\mu y)^{2})}{(\mu x)^{2}} =\mu ^{0}. F(x,y)
Hence it is a homogeneous equation

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}
                                 
Substitute this value in equation (i), we get

\\=v+\frac{xdv}{dx}= -v- v^{2}\\ =\frac{xdv}{dx}=-v(v+2)\\ =\frac{dv}{v+2}=-\frac{dx}{x}\\ =1/2[\frac{1}{v}-\frac{1}{v+2}]dv=-\frac{dx}{x}

Integrating on both sides, we get;

\\=\frac{1}{2}[\log v -\log(v+2)]= -\log x+\log C\\ =\frac{v}{v+2}=(C/x)^{2}

replace the value of v=y/x

\frac{x^{2}y}{y+2x}=C^{2}.............................(ii)

Now y =1 and x = 1

C = 1/\sqrt{3}
therefore, 

\frac{x^{2}y}{y+2x}=1/3

Required solution

Posted by

manish

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