How To Solve A Challenging Differential Equation

[@douglashopkins8070]

Reminds me of the day I discovered differential operators. Made life so much easier.

[@anthonycheng1765]

Can Let u=y' to reduce to second order ODE, first solve y, then integration to solve u

[@holyshit922]

Reduction of order by substitution possible but not necessary After solving homogeneous equation variation of parameters is needed For homogeneous equation we assume that particular solutions are in the form y = e^{λx}

[@giuseppemalaguti435]

Credo che si possa risolvere col metodo della Risposta Impulsiva....che permette di calcolare una yp a partire dall'omogenea ..ma non ricordo... La risposta impulsiva è g(x) =1-cosx..percio la yp(x) =ln[secx+tgx] - xcosx-sinxlnsecx

[@SyberMath]

Hmm. I am not familiar with that

[@giuseppemalaguti435]

​@@SyberMathla risposta impulsiva g(x) é la soluzione dell'equazione omogenea con y(0)=y'(0)=0,y''(0)=1...quindi la yp è l'integrale da 0 a x di g(x-t)* sect dt

[@holyshit922]

Do you want proposition for video Calculate nth derivative of te^(xt)/(e^t - 1) We could use product rule for nth derivative but we have to guess or calculate nth derivative of the factors From my observations d^n/dt^n te^(xt) = x^(n-1)(n+xt)e^(xt) , n > 0 d^n/dt^n 1/(e^t - 1) = (-1)^n*e^t*P_{n-1}(e^{t})/((e^t - 1)^(n+1)) , n > 0 P_{n-1}(x) is polynomial of n-1 degree I observed that polynomial P_{n-1}(x) is palindromic We can make these formulas valid for n >= 0 d^n/dt^n te^(xt) = x^(n-1)(n+xt)e^(xt) , n >= 0 d^n/dt^n 1/(e^t - 1) = (-1)^n*P_{n}(e^{t})/((e^t - 1)^(n+1)) , n >= 0 P_{n}(x) is polynomial of n degree but here P_{n}(x) is no more palindromic If you find formula for P_{n}(x) the rest would be easy

[@viktor-kolyadenko]

In 8:18 we can use: en.wikipedia.org/wiki/Cramer%27s_rule