InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Solve the following differential equation:`((x^2-1)dy)/(dx)+2x y=1/(x^2-1); |x| !=1` |
|
Answer» Given different equation is `(x^(2)-1)(dy)/(dx)+2xy=(1)/(x^(2)-1)` `Rightarrow (dy)/(dx)+((2x)/(x^(2)-1))y=(1)/((x^(2)-1)^(2))` Which is linear differential equation. On comparing it with `(dy)/(dx)+Py=Q`, we get `P=(2x)/(x^(2)-1).Q=(1)/((x^(2)-1)^(2))` `IF=e^(intpdx)=e^(int((2x)/(x^(2)-1))^(dx))` `"Put" x^(2)-1=t Rightarrow 2xdx=dt` `therefore IF=r^(int(dt)/(t))=e^(log t)=t=(x^(2)-1)` The complete solution is `y.IF=int Q.IF+K` `Rightarrow y.(x^(2)-1)=int(1)/((x^(2)-1)^(2)).(x^(2)-1)dx+K` `Rightarrow y.(x^(2)-1)=int(dx)/((x^(2)-1))+K` `Rightarrow y.(x^(2)-1)=(1)/(2)log ((x-1)/(x+1))+K` |
|