Saved Bookmarks
| 1. |
If `y=e^(-x)(A cos x + B sin x)` then y is a situation ofA. `(d^(2)y)/(dx^(2))+2(dy)/(dx)=0`B. `(d^(2)y)/(dx^(2))-2(dy)/(dx)+2y=0`C. `(d^(2)y)/(dx^(2))+2(dy)/(dx)+2y=0`D. `(d^(2)y)/(dx^(2))+2y=0` |
|
Answer» Given that, `y=e^(-x)(Acosx=Bsinx)` On differentiating both sides w.r.t., x we get `" "(dy)/(dx)=-e^(-x)(Acosx+Bcosx)+e^(-x)(-Asinx+Bcosx)` `" "(dy)/(dx)=-y+e^(-x)(-Asinx+Bcosx)` Again, differentiating both sides w.r.t., x, we get `" "(d^(2)y)/(dx^(2))=(-dy)/(dx)+e^(-x)(-cosx-Bsinx)-e^(-x)(-Asin+Bcosx)` `rArr" "(d^(2)y)/(dx^(2))=-(dy)/(dx)-y-[(dy)/(dx)+y]` `rArr" "(d^(2)y)/(dx^(2))=-2(dy)/(dx)-2y` `rArr" "(d^(2)y)/(dx^(2))+2(dy)/(dx)+2y=0` |
|