$\frac{d^4y}{dx^4}-\cot x\frac{d^3y}{dx^3}=0$

Question

Can I solve the DE $\frac{d^4y}{dx^4}-\cot x\frac{d^3y}{dx^3}=0$ as follows?

Complementary function: $m^4-\cot x\times m^3=0\implies m=0,0,0,\cot x$

So $y=a+bx+cx^2+x^3\cot x$

Will it work?

Answer 1

Hint:

If $\dfrac{d^3y}{dx^3}=m$

we have $$\dfrac{dm}{dx}=m\cot x\iff\dfrac{dm}m=\cot x\ dx$$

Integrating both sides we get $\ln m=\ln\sin x+\ln k\iff m=k\sin x$

If $\dfrac{d^2y}{dx^2}=n$

$$\dfrac{dn}{dx}=k\sin x\iff dn=k\sin x\ dx$$

Integrating both sides we get $n=-k\cos x+a$

Then if $\dfrac{dy}{dx}=u,$ $$\dfrac{du}{dx}=-k\cos x+a\iff du=(a-k\sin x)\ dx$$

Integrating both sides we get $$u=ax-k\cos x+b$$

$$\implies dy=(ax-k\cos x+b)dx$$ where $a,b,k$ are arbitrary constants

Now integrate both sides.

Answer 2

Its a reduction of order ODE. Make the substitution $$y'''=p(x)$$

Then $$y^{(IV)}-y'''cot(x)=0\quad \implies\quad p'-p\cot(x)=0$$

Which its solved by either separable or linear method.

$$p=C\sin(x)$$

Replace $p$ and solve by iterative integration: $$\color{red}{y(x)=C_1\cos(x)+C_2x^2+C_3x+C_4}$$

Answer 3

$$\frac{d^4y}{dx^4}-\cot x\frac{d^3y}{dx^3}=0$$ $$\sin x \frac{d^4y}{dx^4}-\cos x\frac{d^3y}{dx^3}=0$$ divide by $sin^2 x$ $$\sin x \frac{d^4y}{dx^4}-\cos x\frac{d^3y}{dx^3}=0$$ $$(\frac{y'''}{\sin x })'=0$$ Integrate $$ \frac{y'''}{ \sin x}=c_1$$ $$ \implies y'''=c_1 \sin x$$ Integrate three times ...to get $y(x)$ Finally $$y(x)=c_1 \cos x +c_2+c_3x+c_4x^2$$