Answer to Question #224866 in Chemical Engineering for Lokika

Question

Solve d³y/dx³-2 d²y/dx²+ dy/dx=sinh.x+2019

Answer 1

Characteristic equation given by

$x^3 - 2 x^2 +x - \sinh x-2019= 0$

Th general solution

$y'''\:-2y''\:+y=0\\ \mathrm{Rewrite\:the\:equation\:with\:}y=e^{γt}\\ \left(\left(e^{γt}\right)\right)'''\:-2\left(\left(e^{γt}\right)\right)''\:+e^{γt}=0\\ e^{γt}\left(γ^3-2γ^2+1\right)=0\\ γ=1,\:γ=\frac{1+\sqrt{5}}{2},\:γ=\frac{1-\sqrt{5}}{2}\\ \mathrm{For\:non\:repeated\:real\:roots\:}γ_1,\:γ_2,\:...,\:γ_n\mathrm{,\:the\:general\:solution\:takes\:the\:form:}\\ y=c_1e^{γ_1\:t}+c_2e^{γ_2\:t}+...+c_ne^{γ_n\:t}\\ c_1e^t+c_2e^{\frac{1+\sqrt{5}}{2}t}+c_3e^{\frac{1-\sqrt{5}}{2}t}\\ y=c_1e^t+c_2e^{\frac{\left(1+\sqrt{5}\right)t}{2}}+c_3e^{\frac{\left(1-\sqrt{5}\right)t}{2}}\\$

Particular solution

$y_p=\frac{\sinh t +2019}{D^3-2D^2+D}\\ =\frac{1}{2}[1+t-\frac{1}{6}(\sinh t)]\\ \therefore y= c_1e^t+c_2e^{\frac{\left(1+\sqrt{5}\right)t}{2}}+c_3e^{\frac{\left(1-\sqrt{5}\right)t}{2}}+\frac{1}{2}[1+t-\frac{1}{6}(\sinh t)]\\$