# transcendental DE

#### idontknow

(1) $$\displaystyle yy'y''-4x^3=0.$$

(2) $$\displaystyle y^{(4)}(x) -y(x)=k.$$

#### SDK

Neither of those is transcendental. What exactly are you asking?

topsquark

#### topsquark

Math Team
(1) Noting that $$\displaystyle y y' y'' = 4x^3$$ I looked for a polynomial solution. I got $$\displaystyle y = x^2$$.

(2) Break this up into homogeneous and particular parts. So solve $$\displaystyle y_h^{(4)} - y_h = 0$$. The characteristic equation is $$\displaystyle m^4 - 1 = 0$$ which has solutions of m = $$\displaystyle \{ \pm 1, ~ \pm i \}$$. The homogeneous solution is therefore $$\displaystyle y_h(x) = A e^{x} + B e^{-x} + C ~ \sin(x) + D ~ \cos(x)$$.

Now find the particular solution...

-Dan

idontknow