Question

Help :)
solve this using transformation of Laplace ..
y(0)=1 and y'(0)=0

Answer 1

just apply laplace to both sides of each equation and enjoy

Answer 2

for example number 4
after applying laplace transform u have

\[s^2Y(s)-sy(0)-y'(0)-2Y(s)=0\]\[Y(s)(s^2-2)=s\]\[Y(s)=\frac{s}{s^2-2}\]

use inverse laplace to get \(y\) because we know that \(L^{-1} (Y)=y\)

Answer 3

u can do the rest easily...

Answer 4

i dont understand , can u give me another example ?

Answer 5

anyone ?

Answer 6

i will solve no 3
taking laplace
since
\[L (y'')=(s^2Y-sy(0)-y'(0))\]
and Laplace of
\[L (y')=(sY-y(0))\]
so put in the equation
\[\Large x[s^2Y-sy(0)-y'(0)]+[sY-y(0)]+x=0\]
using values of y'(0)=0 ,y(0)=1
\[\Large x[s^2Y-s-0]+[sY-1]+x=0\]
\[\Large x[s^2Y-s]+[sY-1]+x=0\]
now since
\[\Large L(xf(x))=(-1) \frac{d}{ds}(L(f(x))\]
also
\[\Large L(x)=\frac{1}{s^2}\]
so it becomes
\[\Large \frac{d}{ds}[s^2Y-s]+[sY-1]+\frac{1}{s^2}=0\]
\[\Large 2sY-s^2\frac{dY}{ds}-1+sY-1+\frac{1}{s^2}=0\]
(used the product rule of differentiation above)
simplify
\[\Large (2s+s)Y-s^2\frac{dY}{ds}-2+\frac{1}{s^2}=0\]
\[\Large -s^2\frac{dY}{ds}+(2s+s)Y=2-\frac{1}{s^2}\]
divide by -s^2
\[\Large \frac{dY}{ds}-\frac{3s}{s^2}Y=\frac{1}{s^4}-\frac{2}{s^2}\]
\[\Large \frac{dY}{ds}-\frac{3}{s}Y=\frac{1}{s^4}-\frac{2}{s^2}\]
it is Linear first order equation .you will use integration factor Method here.
can you solve this equation ??

Answer 7

\[-2sY-s^2 \frac{dY}{ds}+1\] 5th line from bot ...check it again

Answer 8

oopss .. i did it again . one typo
i missed the minus sign :(((((((((((((((((((((

Answer 9

this minus changed the whole question :(

Answer 10

let me try again .

Answer 11

continue from above
also
\[\Large L(x)=\frac{1}{s^2}\]
it becomes
\[\Large -\frac{d}{ds}[s^2Y-s]+[sY-1]+\frac{1}{s^2}=0\]
use product rule.
\[\Large -2sY-s^2\frac{dY}{ds}+1+sY-1+\frac{1}{s^2}=0\]
simplify
\[\Large -s^2\frac{dY}{ds}-sY=-\frac{1}{s^2}\]
divide by -s^2
\[\Large \frac{dY}{ds}+\frac{Y}{s}=\frac{1}{s^4}\]
i hope you can solve this Linear differential equation.