Question

The differential equation which has y 4 = Cx3 − 3 as its general solution is:

Answer 1

$y^ 4 = Cx^3 − 3$ ...(1)

Differentiating both sides w.r.t $x$ , we get

$4y^3y'=3Cx^2\Rightarrow C=\frac {4y^3y'}{3x^2}$

Put the value of C in equation (1), we get

$y^4=\frac {4y^3y'}{3x^2}\times x^3-3=\frac {4xy^3y'}{3}-3=\frac {4xy^3y'-9}{3}\\ 3y^4-4xy^3y'+9=0$

is the differential equation.