Question #189062

X^3=a+bp


1
Expert's answer
2021-05-07T12:31:38-0400

Ans:-

x3=a+bpx^3=a+bp

differentiate with respect to p

3x2dxdp=b3x^2\dfrac{dx}{dp}=b

again differentiate with respect to p

6x(dxdp)2+3x2d2xdp2=06x(\dfrac{dx}{dp})^2+3x^2\dfrac{d^2x}{dp^2}=0

6x(2(dxdp)2+xd2xdp2)=06x(2(\dfrac{dx}{dp})^2+x\dfrac{d^2x}{dp^2})=0

xd2xdp2+2(dxdp)2+=0\Rightarrow x\dfrac{d^2x}{dp^2}+ 2(\dfrac{dx}{dp})^2+=0

This is required Differential equation


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