अवकल समीकरण `x(1-x^(2))dy+(2x^(2)y-y-5x^(3))dx=0` का हल है

Options

A. `y=x+Cxsqrt(1-x^(2))`
B. `5y=x+Csqrt(1-x^(2))`
C. `y=5x+Cxsqrt(1-x^(2))`
D. इनमे से कोई नहीं

Answer

C

Explanation

`because x(1-x^(2))dy+(2x^(2)y-y-5x^(3))dx=0`
`impliesx(1-x^(2))(dy)/(dx)+(2x^(2)-1)y=5x^(3)`
`implies(dy)/(dx)+(2x^(2)-1)/(x(1-x^(2)))y=(5x^(3))/(x(1-x^(2)))`
स्पष्टत: यह रैखिक अवकल समीकरण है। यहाँ, `P=(2x^(2)-1)/(x(1-x^(2)))` और `Q=(5x^(2))/(1-x^(2))`
`therefore int Pdx=int(2x^(2)-1)/(x(1-x^(2)))dx=int((2x^(2)-1)x)/(x^(2)(1-x^(2)))dx`
माना `x^(2)=timplies2xdx=dt`
`=(1)/(2)int((2t-1))/(t(1-t))dt=(1)/(2)int(t-(1-t))/(t(1-t))dt`
`=(1)/(2)int((1)/(1-t)-(1)/(t))dt`
`=(1)/(2)[-log(1-t)-logt]`
`=-(1)/(2)logt(1-t)`
`=-(1)/(2)logx^(2)(1-x^(2))=log.(1)/(xsqrt(1-x^(2)))`
`therefore IF =e^(intPdx)=e^(log.(1)/(xsqrt(1-x^(2))))=(1)/(xsqrt(1-x^(2)))`
अतः अभीष्ट हल है,
`y(IF)=intQ(IF)dx+C`
`y.(1)/(xsqrt(1-x^(2)))=int(5x^(2))/(1-x^(2)).(1)/(xsqrt(1-x^(2)))dx+C`
`=5int(x)/((1-x^(2))^(3//2))dx+C`
`1-x^(2)=timplies-2xdx=dt=-5.(1)/(2)int(dt)/(t^(3//2))+C=-(5)/(2)(-2)(1)/(sqrt(t))+C`
`impliesy.(1)/(xsqrt(1-x^(2)))=(5)/(sqrt(1-x^(2)))+C`
`therefore y=5x+Cxsqrt(1-x^(2))`