Question

Find the general solutions of the following differential equations using D-operator methods:

2.1 (D² - 2D + 5)y = x+5

2.2 (D+4)² x = sinh4t

Answer 1

$2.1:\\\left( D^2-2D+5 \right) y=x+5\\\phi \left( D \right) =D^2-2D+5=\left( D-1 \right) ^2+2^2\\\left( \left( D-1 \right) ^2+2^2 \right) y=x+5\\The\,\,solution\,\,of\,\,the\,\,homogeneous\,\,equation\,\,is\\y=Ae^x\cos 2x+Be^x\sin 2x\\Particular\,\,solution:\\y=C_0+C_1x:\\\left( D^2-2D+5 \right) \left( C_0+C_1x \right) =x+5\\-2C_1+5\left( C_0+C_1x \right) =x+5\\\left\{ \begin{array}{c} -2C_1+5C_0=5\\ 5C_1=1\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} C_1=0.2\\ C_0=1.5\\\end{array} \right. \\y=1.5+0.2x+Ae^x\cos 2x+Be^x\sin 2x\\2.2:\\\left( D+4 \right) ^2x=\sinh 4t\\e^{-4t}D^2e^{4t}x=\sinh 4t\\D^2e^{4t}x=e^{4t}\sinh 4t\\D^2e^{4t}x=\frac{1}{2}e^{8t}-\frac{1}{2}\\De^{4t}x=\int{\left( \frac{1}{2}e^{8t}-\frac{1}{2} \right) dt}=\frac{1}{16}e^{8t}-\frac{1}{2}t+C_1\\e^{4t}x=\int{\left( \frac{1}{16}e^{8t}-\frac{1}{2}t+C_1 \right) dt}=\frac{1}{128}e^{8t}-\frac{1}{2t^2}+C_1t+C_2\\x=\frac{1}{128}e^{4t}-\frac{1}{2t^2}e^{-4t}+C_1te^{-4t}+C_2e^{-4t}$