Cauchy problem: $x'-cot(t)\cdot x =2tsint; x\left(\frac{\pi}{2}\right)=\frac{\pi^2}{4}$

27 Views Asked by At

Here is the problem: $x'-cot(t)\cdot x =2tsint; x\left(\frac{\pi}{2}\right)=\frac{\pi^2}{4}$

My solution: $$e^{\int -cot(t)dt}=e^{-ln(|sint|}=-sint$$

$$x'\cdot (-sin(t)) -cot(t) \cdot (-sin(t)) \cdot x=2tsin^2t$$

$$x' \cdot (-sin(t)) + cos(t) \cdot x=2tsin^2t$$

So here already is a problem, because deriative of $-sin(t)$ is $-cos(t)$. Where is my error?

1

There are 1 best solutions below

0
On BEST ANSWER

$$x'-\cot(t)\cdot x =2t\sin(t)\quad ; \quad x\left(\frac{\pi}{2}\right)=\frac{\pi^2}{4}$$ $$e^{\int -\cot(t)dt}=e^{-\ln(|\sin(t)|)}=\frac{1}{|\sin(t)|}\quad\text{but not}\quad=\sin(t)$$ What you wrote is not detailed enough to see where exactly is your difficulty. Especially where the condition $x\left(\frac{\pi}{2}\right)=\frac{\pi^2}{4}$ is used in your calculus ?

Alternatively the method of "variation of parameter"consists first to solve the homogeneous part of the ODE : $x'-\cot(t)\cdot x =0$ which is separable and leads to $x=C\:\sin(t)$ then to replace the parameter $C$ by a function $f(t)$ which leads to : $$x'=f'\sin(t)+f\cos(t)-\cot(t)f\sin(t)=2t\sin(t)$$ $$f'\sin(t)=2t\sin(t)\quad\implies\quad f=t^2+c$$ $$x=t^2\sin(t)+c\sin(t)$$ $$x\left(\frac{\pi}{2}\right)=\frac{\pi^2}{4}= (\pi/2)^2\sin(\pi/2)+c\sin(\pi/2)=\frac{\pi^2}{4}+c\quad\implies\quad c=0$$ $$x(t)=t^2\sin(t)$$