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d-2-y-dx-2-cy-x-2-y-x-




Question Number 1964 by 123456 last updated on 26/Oct/15
(d^2 y/dx^2 )=((cy)/x^2 )  y(x)=?
$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{cy}}{{x}^{\mathrm{2}} } \\ $$$${y}\left({x}\right)=? \\ $$
Answered by prakash jain last updated on 26/Oct/15
y=kx^n   y′′=kn(n−1)x^(n−2) =ckx^(n−2)   n(n−1)=c  n^2 −n−c=0  n=((1±(√(1+4c)))/2)  y=c_1 x^((1+(√(1+4c)))/2) +c_2 x^((1−(√(1+4c)))/2)
$${y}={kx}^{{n}} \\ $$$${y}''={kn}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} ={ckx}^{{n}−\mathrm{2}} \\ $$$${n}\left({n}−\mathrm{1}\right)={c} \\ $$$${n}^{\mathrm{2}} −{n}−{c}=\mathrm{0} \\ $$$${n}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4c}}}{\mathrm{2}} \\ $$$${y}={c}_{\mathrm{1}} {x}^{\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{c}}}{\mathrm{2}}} +{c}_{\mathrm{2}} {x}^{\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}{c}}}{\mathrm{2}}} \\ $$

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