Question

Find the initial value problem of y'=y+x/y-x, y(0)= 2

Answer 1

$\text{We set y = vx}\implies \frac{dy}{dx}= v+x\frac{dy}{dx}$

$\implies v+x\frac{dy}{dx}= \frac{vx+x}{vx-x}= \frac{v+1}{v-1} \\ \implies x\frac{dy}{dx} = -\frac{v^2-2v-1}{v-1}\\-\int\frac{v-1}{v^2-2v-1}=\int\frac{dx}{x}\\=-\frac{1}{2}In(v^2-2v-1)=Inx+inA$

$=In(v^2-2v-1)=-2InAx\\\text{recall that y = vx}\implies v=\frac{y}{x}\\\implies In(\frac{y^2}{x^2}-\frac{2y}{x}-1) =-2InAx\\=In(y^2-2xy-x^2) - Inx^2 = -2InAx\\=In(y^2-2xy-x^2)= In\frac{x^2}{A^2x^2}\\=In(y^2-2xy-x^2)= In\frac{1}{A^2} = inD \text{, by setting D = } \frac{1}{A^2}\\\text{Therefore, }y^2-2xy-x^2=D, \text{Given y(0) = 2}\\\text{Hence, }D=4\implies y^2 -2xy-x^2=4$