Answer:
Required solution is [tex]y(x)=1+A\cos x+B\sin x[/tex] where A and B are constants.
Step-by-step explanation:
Given nonhomogenous differential equation is,
[tex]y''(x)+y'(x)=1\hfill (1)[/tex] with [tex]y_1(x)=1[/tex]
To find another solution, consider [tex]m=\frac{\partial}{\partial x}[/tex] be such that,
[tex]m^2+1=0\implies m=\pm i[/tex]
Hence,
[tex]C.F=A\cos x+B\sin x[/tex] where A and B are constants.
Let [tex]D=\frac{\partial}{\partial x}[/tex]
[tex]P.I=\frac{1}{1+D^2}(1)[/tex]
[tex]=(1+D^2)^{-1}(1)[/tex]
[tex]=(1-D^2+.....)(1)=1[/tex]
Hence,
[tex]y(x)=1+A\cos x+B\sin x[/tex] where A and B are constants.
which is required solution.
