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  • Solve for particular solution. (x + y) dy + (x - y) dx = 0; y = 1 when x = 1

Solve for particular solution.

    Q11.    (x + y)dy + (x -y)dx = 0;\ y =1\ when \ x =1

Answers (1)

best_answer

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

We can clearly say that 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)

\\v+x\frac{dv}{dx}=\frac{v-1}{v+1}\\ \Rightarrow x\frac{dv}{dx} = -\frac{(1+v^{2})}{1+v}

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

On integrating both sides

\\=\frac{1}{2}[\log (1+v^{2})]+\tan^{-1}v = -\log x +k\\ =\log(1+v^{2})+2\tan^{-1}v=-2\log x +2k\\ =\log[(1+(y/x)^{2}).x^{2}]+2\tan^{-1}(y/x)=2k\\ =\log(x^{2}+y^{2})+2\tan^{-1}(y/x) = 2k......................(ii)

Now, y=1 and x= 1


\\=\log 2 +2\tan^{-1}1=2k\\ =\pi/2+\log 2 = 2k\\

After substituting the value of 2k in eq. (ii)

\log(x^{2}+y^2)+2\tan^{-1}(y/x)=\pi/2+\log 2

This is the required solution.

Posted by

manish

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