Answer to Question #181880 in Differential Equations for Keiran

Question

y'y'''-(y'')^2=(y'')^3

Answer 1

Solution :-

y'y'''-(y'')^2=(y'')^3

= $({dy\over dx})({d^3y\over dx^3})-({d^2y\over dx^2})^2=({d^2y\over dx^2})^3$

=$({dy\over dx})({d^3y\over dx^3})-({d^2y\over dx^2})^2-({d^2y\over dx^2})^3 = 0$ ........(1)

manuplating the (1) equation

$f\left( x \right) \cdot {\large\frac{d}{{dx}}\normalsize} \left[ {\ln f\left( x \right)} \right]$ = $\frac {dy}{dx}$ .....(2)

= $\intop\intop\intop F(y)dx$

x = log(y)

using the properties and (1) , (2)

y = e^x

=$\intop\intop C_1e^{xc_1}.e^{c_1}$ dx.dx

=$\intop e^{xc_1}.e^{c_1c_2}$ dx

answer = $\boxed{y(x)={e^{xc_1}e^{c_1c_2}\over c_1} + C_3}$