Question

w''+w=t^2+2 w(0)=1 w'(0)=-1
laplace transform ivp's
not sure what i need to do

Answer 1

i tried working it but laplace of w'' gives
\[s^2W(s)-1s+1+W(s)\]
which isn't the same as our starting equation

Answer 2

@oldrin.bataku if you could help with this when you get a chance it would be appreciated!

Answer 3

great and all my next problems have to be 'time shifted' if you don't mind @oldrin.bataku i will need help with those too. unless the substitution used is just whatever the x value of the IVP is. From my example in class its a little confusing as we have
\[w''-2w'+w=6t+2\]
with initial conditions of
\[w(-1)=3\] and \[w'(-1)=7\]
we used \[y(t)=w(t-1), y'(t)=w'(t-1), y''(t)=w''(t-1)\]
im guessing we used t-1 because of our initial but not sure

Answer 4

$$w''+w=t^2+2\qquad w(0)=1,w'(0)=-1$$taking the Laplace transform of our equation:$$s^2W-sw(0)-w'(0)+W=\frac2{s^3}+\frac2s\\s^2W+W=\frac2{s^3}+\frac2s+s-1\\(s^2+1)W(s)=\frac2{s^3}+\frac2s+s-1\\(s^2+1)W(s)=\frac{s^4-s^3+2s^2+2}{s^3}\\W(s)=\frac{s^4-s^3+2s^2+2}{s^3(s^2+1)}=\frac{As^2+Bs+C}{s^3}+\frac{Ds+E}{s^2+1}\\s^4-s^3+2s^2+2=(As^2+Bs+C)(s^2+1)+s^3(Ds+E)\\s^4-s^3+2s^2+2=(A+D)s^4+(B+E)s^3+(A+C)s^2+Bs+C\\A+D=1\\B+E=-1\\A+C=2\\B=0\\C=2$$so we see \(A=0,B=0,C=2,D=1,E=-1\) yielding:$$W(s)=\frac2{s^3}+\frac{s}{s^2+1}-\frac1{s^2+1}\\w(t)=t^2+\cos t-\sin t$$

Answer 5

for time shifting since our IVP gives \(w(-1)\) and we want it to be \(y(0)\) we just make \(y(t)=w(t-1)\)