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# Show that 2y1y3=3y2^2 where y1=dy/dx,y2=d^2y/dx^2 and y3=d^3y/dx^3

Vikas TU
14149 Points
3 years ago
To prove that 2y1y3=3y2^2 there should be some function specified for y.Which is not given here.

Considering Y=(ax+b) / (cx+d)

We can prove the equation 2Y⁽¹⁾Y⁽³⁾= 3[Y⁽²⁾]², where Y = (ax + b)/(cx + d) and  Y⁽¹⁾, Y⁽²⁾, and Y⁽³⁾are the 1st, 2nd, and 3rd order derivatives of function Y, respectively, as follows:

Since the function Y is defined by a rational expression in x: Y(x) = (ax + b)/(cx + d), we’ll find derivatives Y⁽¹⁾,  Y⁽²⁾, and Y⁽³⁾by utilizing the quotient rule of differentiation as follows:

Y⁽¹⁾ = d [(ax + b)/(cx + d)] /dx

= (ad ‒ bc)/(cx + d)²

Y⁽²⁾ = d[Y⁽¹⁾]/dx

= d[(ad ‒ bc)/(cx + d)²]/dx

= ‒ [(2c)(ad ‒ bc)]/(cx + d)³

Y⁽³⁾ = d[Y⁽²⁾]/dx

= d[‒ (2c)(ad ‒ bc)/(cx + d)³]/dx

Now, using the expressions for Y⁽¹⁾,  Y⁽²⁾, and Y⁽³⁾to  prove the equation 2Y⁽¹⁾Y⁽³⁾= 3[Y⁽²⁾]² as follows:

2Y⁽¹⁾Y⁽³⁾= 3[Y⁽²⁾]²

Now, substituting for Y⁽¹⁾, Y⁽²⁾, and Y⁽³⁾, we have: