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Q.Solve: (1+x2)dydx=2xy−y2(1+x^2)\dfrac{dy}{dx}=2xy-y^2

Odisha ChseOdisha CHSE +2 Science Board Exam 2022Subjective· 5mImportance★★★★★
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This is a Bernoulli equation in yy (power 2); substitute v=1/yv=1/y to turn it into a linear first-order equation in vv, then solve with an integrating factor.

Given (1+x2)dydx=2xy−y2(1+x^2)\dfrac{dy}{dx}=2xy-y^2.

Divide throughout by y2y^2 (Bernoulli form, n=2n=2):

(1+x2)1y2dydx−2xy=−1(1+x^2)\dfrac{1}{y^2}\dfrac{dy}{dx}-\dfrac{2x}{y}=-1

Let v=1yv=\dfrac1y, so dvdx=−1y2dydx\dfrac{dv}{dx}=-\dfrac{1}{y^2}\dfrac{dy}{dx}, i.e. 1y2dydx=−dvdx\dfrac{1}{y^2}\dfrac{dy}{dx}=-\dfrac{dv}{dx}.

Substitute:

(1+x2)(−dvdx)−2xv=−1(1+x^2)\left(-\dfrac{dv}{dx}\right)-2xv=-1

−(1+x2)dvdx−2xv=−1-(1+x^2)\dfrac{dv}{dx}-2xv=-1

(1+x2)dvdx+2xv=1(1+x^2)\dfrac{dv}{dx}+2xv=1

Divide by (1+x2)(1+x^2):

dvdx+2x1+x2v=11+x2\dfrac{dv}{dx}+\dfrac{2x}{1+x^2}v=\dfrac{1}{1+x^2}

This is linear in vv. Integrating factor:

I.F.=e∫2x1+x2dx=eln⁡(1+x2)=1+x2\text{I.F.}=e^{\int\frac{2x}{1+x^2}dx}=e^{\ln(1+x^2)}=1+x^2.

Multiply through by the I.F. and integrate:

ddx[v(1+x2)]=(1+x2)⋅11+x2=1\dfrac{d}{dx}\big[v(1+x^2)\big]=(1+x^2)\cdot\dfrac{1}{1+x^2}=1

v(1+x2)=x+Cv(1+x^2)=x+C

Substitute back v=1/yv=1/y:

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