Hey MHF,
I got stuck on finding the gen sol. to the following ODE
$\displaystyle y^{(4)}+4y = 0$
I solved the auxiliary eqn. $\displaystyle m^4+4= 0$ for $\displaystyle m= ^4\sqrt{-4}$
So the roots must be complex? This is where I got stuck.
Hey MHF,
I got stuck on finding the gen sol. to the following ODE
$\displaystyle y^{(4)}+4y = 0$
I solved the auxiliary eqn. $\displaystyle m^4+4= 0$ for $\displaystyle m= ^4\sqrt{-4}$
So the roots must be complex? This is where I got stuck.
Ok so I can use that to solve the IVP.
$\displaystyle y(0)=0$ so $\displaystyle 0 = C1 + C3$
and
$\displaystyle y'(x)= C1e^x(cosx-sinx)+C2e^x(sinx+cosx)-C3e^{-x}$$\displaystyle (cosx+sinx)+C4e^{-x}(cosx-sinx)$
$\displaystyle y'(0) = 1$ so
$\displaystyle 1=C1+C2-C3+C4$
Is this correct so far?
If what I did above is correct then
$\displaystyle y''(x)= 2C2e^xcosx-2C1e^xsinx+2C3e^{-x}sinx-2C4e^{-x}cosx$ then
$\displaystyle 0=2C2-2C4$
$\displaystyle y'''(x)= 2C2e^x(cosx-sinx)-2C1e^x(sinx+cosx)+2C3e^{-x}$$\displaystyle (cosx-sinx)-2C4e^{-x}(cosx+sinx)$ and
$\displaystyle 2= 2C2-2C1+2C3-2C4$
Idk if i'm on the right track here...
$\displaystyle y(x) = C_{1} e^x \cos x + C_{2} e^{x} \sin x + C_{3} e^{-x} \cos x + C_{4}e^{-x} \sin x $
$\displaystyle y'(x) = C_{1} (e^{x} \cos x - e^{x} \sin x) + C_{2} (e^{x} \sin x + e^{x} \cos x) + C_{3} (-e^{-x} \cos x - e^{-x} \sin x) + C_{4} $ $\displaystyle (-e^{-x} \sin x+ e^{-x} \cos x) $
$\displaystyle y''(x) = C_{1}(e^{x} \cos x -e^{x} \sin x -e^{x} \sin x - e^{x} \ \cos x) + C_{2}( e^{x}\sin x + e^{x}\cos x + e^{x} \cos x - e^{x} $ $\displaystyle \sin x) + C_{3} (e^{-x}\cos x + e^{-x} \sin x+ e^{-x} \sin x - e^{-x} \cos x) + C_{4} $ $\displaystyle ( e^{-x} \sin x - e^{-x} \cos x -e^{-x} \cos x - e^{-x} \sin x)$
$\displaystyle = -2C_{1}(e^{x}\sin x) + 2 C_{2}e^{x}\cos x + 2C_{3}e^{-x}\sin x -2C_{4} e^{-x} \cos x $
$\displaystyle y'''(x) = -2C_{1}(e^{x} \sin x + e^{x} \cos x) + 2C_{2}(e^{x} \cos x - e^{x} \sin x) + 2C_{3} (-e^{-x} \sin x +e^{-x} \cos x) $ $\displaystyle - 2C_{4} (-e^{-x}\cos x - e^{-x} \sin x) $
$\displaystyle y(0) = 0 = C_{1}+ C_{3} $
$\displaystyle y'(0) = 1= C_{1} + C_{2} - C_{3} + C_{4} $
EDIT: $\displaystyle y''(0) = 0 = 2C_{2} - 2C_{4} $
$\displaystyle y'''(0) = 2 = -2 C_{1} + 2C_{2} +2C_{3} + 2C_{4} $
which when solved, gives $\displaystyle C_{1}=0, C_{2} = \frac{1}{2}, C_{3} =0, C_{4} = \frac{1}{2} $
Your y'' matches the one I have above. However isn't sin(0)=0? So the C1 and C3 terms would go away when evaluated at 0? Leaving:
$\displaystyle y''(0)= C1(-2e^0(sin(0))+C2(2e^0(cos(0)))+C3(2e^0(sin(0)))$
$\displaystyle +C4(-2e^0(cos(0)))$
$\displaystyle =>0=0+2C2+0-2C4$
?
Ok so I had the last sign on y''' wrong yours was correct though. So we have
$\displaystyle y(0)= 0 =C1+C3$
$\displaystyle y'(0)= 1 =C1+C2-C3+C4$
$\displaystyle y''(0)= 0 =2C2-2C4$
$\displaystyle y'''(0)= 2 =-2C1+2C2+2C3+2C4$
How do you deduct the constants from these? The only thing I could see is that
$\displaystyle C2=C4$
From the first equation we also know that $\displaystyle C_{1}=-C_{3}$
Then from the second equation $\displaystyle 2C_{1}+2C_{2} =1 $ (1)
And from the fourth equation $\displaystyle -4C_{1}+4C_{2}=2 $ or $\displaystyle -2C_{1}+2C_{2}=1 $ (2)
Add (1) and (2) to get $\displaystyle 4C_{2} = 2 $
so $\displaystyle C_{2} = \frac{1}{2}$ which means that $\displaystyle C_{4}=\frac{1}{2} $
and going back to (1), $\displaystyle 2C_{1}+2\Big(\frac{1}{2}\Big) = 1 $
so $\displaystyle C_{1} = 0$ which means that $\displaystyle C_{3}=0 $