Respuesta :
Answer:
[tex]y_1, y_2[/tex] satisfies given homogenous solution and the particular solution is [tex]\frac{5}{t^4}+\frac{1}{2t^2}-\frac{5}{2}[/tex].
Step-by-step explanation:
Given homogeneous equation is,
[tex]t^2y''-2y=5t^2-1[/tex] where t>0 subject to [tex]y_1(t)=t^2, y_2(t)=\frac{1}{t}\hfill (1)[/tex]
To verify,
- wheather [tex]y_1, y_2[/tex] satisfies given homogenous equation, then both will satisfy [tex]t^2y''-2y=0[/tex] that is,
[tex]t^2y_{1}^{''}-2y_1=t^2\times 2-2\times t^2=0[/tex]
[tex]t^2y_{2}^{''}-2y_2=t^2\times 2t^{-3}-2\times t^{-1}=2t^{-1}-2t^{-1}=0[/tex]
Thus [tex]y_1[/tex] and [tex]y_2[/tex] satisfies (1).
Now the wronskean,
[tex]W(y_1, y_2)(x)=y_1y_{2}^{'}-y_{2}y_{1}^{'}=-t^2t^{-2}-2tt^{-1}=-3\neq 0[/tex]
Thus [tex]y_1[/tex] and [tex]y_2[/tex] are solution of (1) .
- Particular solution,
[tex]Y(t)=\frac{1}{t^2D^2-2}(5t^2-1)[/tex] where [tex]D\equiv \frac{\partial }{\partial t}[/tex]
[tex]=-\frac{1}{2t^2}\frac{1}{1-\frac{D^2}{t^2}}()5t^2-1[/tex]
[tex]=-\frac{1}{2t^2}(1-\frac{D^2}{t^2})^{-1}(5t^2-1)[/tex]
[tex]=-\frac{1}{2t^2}(1-\frac{D^2}{t^2}+........)(5t^2-1)[/tex]
[tex]=-\frac{1}{2t^2}(5t^2-1-\frac{10}{t^2})[/tex] sincce [tex]D^2(5t^2-1)=10[/tex]
[tex]=\frac{5}{t^4}+\frac{1}{2t^2}-\frac{5}{2}[/tex]
Hence the result.
