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Question 87 of 126

Q.Solve : (x2+y2) dx+3xy dy=0(x^2+y^2)\,dx + 3xy\,dy = 0

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2016Subjective· 10mImportance★★★★★
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Recognise (x2+y2)dx+3xy dy=0(x^2+y^2)dx+3xy\,dy=0 as homogeneous of degree 2, substitute y=vxy=vx, separate variables, and integrate.

  1. Check homogeneity. Write as dydx=−x2+y23xy\dfrac{dy}{dx}=-\dfrac{x^2+y^2}{3xy}. Both numerator and denominator of the right side are homogeneous of degree 2 in x,yx,y, so the equation is homogeneous.

  2. Substitute y=vxy=vx. Then dydx=v+xdvdx\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}, i.e. dy=v dx+x dvdy=v\,dx+x\,dv.

    Substitute into (x2+y2)dx+3xy dy=0(x^2+y^2)dx+3xy\,dy=0:

    (x2+v2x2)dx+3x(vx)(v dx+x dv)=0(x^2+v^2x^2)dx+3x(vx)(v\,dx+x\,dv)=0

    x2(1+v2)dx+3vx2(v dx+x dv)=0x^2(1+v^2)dx+3vx^2(v\,dx+x\,dv)=0

  3. Divide by x2x^2 and expand:

    (1+v2)dx+3v2 dx+3vx dv=0(1+v^2)dx+3v^2\,dx+3vx\,dv=0

    (1+4v2)dx+3vx dv=0(1+4v^2)dx+3vx\,dv=0

  4. Separate variables.

    dxx=−3v1+4v2 dv\dfrac{dx}{x}=-\dfrac{3v}{1+4v^2}\,dv

  5. Integrate both sides. For the right side, let w=1+4v2w=1+4v^2, dw=8v dvdw=8v\,dv, so v dv=dw8v\,dv=\dfrac{dw}{8}:

    ∫−3v1+4v2dv=−38∫dww=−38ln⁡∣1+4v2∣\displaystyle\int-\dfrac{3v}{1+4v^2}dv=-\dfrac{3}{8}\displaystyle\int\dfrac{dw}{w}=-\dfrac{3}{8}\ln|1+4v^2|

    So: ln⁡∣x∣=−38ln⁡(1+4v2)+C1\ln|x|=-\dfrac{3}{8}\ln(1+4v^2)+C_1

  6. Clear fractions. Multiply through by 88:

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