X^3=a+bp
Ans:-
x3=a+bpx^3=a+bpx3=a+bp
differentiate with respect to p
3x2dxdp=b3x^2\dfrac{dx}{dp}=b3x2dpdx=b
again differentiate with respect to p
6x(dxdp)2+3x2d2xdp2=06x(\dfrac{dx}{dp})^2+3x^2\dfrac{d^2x}{dp^2}=06x(dpdx)2+3x2dp2d2x=0
6x(2(dxdp)2+xd2xdp2)=06x(2(\dfrac{dx}{dp})^2+x\dfrac{d^2x}{dp^2})=06x(2(dpdx)2+xdp2d2x)=0
⇒xd2xdp2+2(dxdp)2+=0\Rightarrow x\dfrac{d^2x}{dp^2}+ 2(\dfrac{dx}{dp})^2+=0⇒xdp2d2x+2(dpdx)2+=0
This is required Differential equation
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