I am trying to solve an IVP problem using the Fourier Transform method. I get stuck halfway through when I reduce the problem down to an ODE.
We have,
$$\frac{d^2u}{dt^2} = 4\frac{d^2u}{dx^2} + (2(2+2t^2+x^2+t^2) \times e^{\frac{-x}{4}}) \tag{1}$$ Where the growth condition is given by,
$$u(x,t), \frac{du}{dx} (x,t) --> 0 \tag{2}$$
And the initial conditions are given by,
$$u(x,0) = 0, \frac{du}{dt}(x,0) = 0 \tag{3}$$
Where $-\infty < x <\infty$ and $t>0$
For solving the above I appeal to the method of Fourier transform.
$$F[u(x,t)](\omega, t) = \Omega(\omega,t) = \int_{-\infty}^\infty u(x,t) \times e^{-i\omega t} dx$$
Then our first term in equation (1),
$$F[\frac{d^2u}{dt^2}(x,t)](\omega, t) = \int_{-\infty}^\infty \frac{d^2u}{dt^2} \times e^{-i\omega x} dx$$
$$=\frac{d^2}{dt^2} \int_{-\infty}^\infty u(x,t) \times e^{-i\omega x} dx = \frac{d^2}{dt^2} \Omega (\omega, t)$$
Then the second term in (1),
$$F[\frac{d^2u}{dx^2} (x,t) (\omega, t) = \int_{-\infty}^\infty \frac{d^2u}{dx^2} (x,t)(\omega, t) \times e^{-i \omega x} dx$$
Because this term involved x, and we are integrating with respect to x, we must integrate this, and we can do this by doing so by parts twice, thus,
$$[\frac{du}{dx} (x,t) \times e^{-i \omega x}]_{-\infty}^\infty + iw \int_{-\infty}^\infty \frac{du}{dx} e^{i\omega x}$$
By (2) $[\frac{du}{dx} (x,t) \times e^{-i \omega x}]_{-\infty}^\infty$ becomes $0$
So we have,
$$0+i \omega ([\frac{du}{dx} (x,t) \times e^{-i \omega x}]_{-\infty}^\infty + i \omega \int_{-\infty}^\infty u(x,t) e^{-i \omega x}$$
By (2) again $([\frac{du}{dx} (x,t) \times e^{-i \omega x}]_{-\infty}^\infty$ goes to $0$
So we have,
$$(i \omega)^2 \Omega(\omega, t) = - \omega^2 \Omega( \omega, t)$$
Therefore,
$$F[\frac{d^2u}{dt^2} - 4\frac{d^2u}{dx^2}](\omega, t) = \Omega''(\omega, t) + 4 \omega^2 \Omega (\omega, t)$$
Next we calculate the Fourier Transform of the non-homogeneus part,
$$F[2(2+2t^2+x^2 t^2)e^{\frac{-x^2}{4}}](\omega, t) = 4F[ e^{\frac{-x^2}{4}}](\omega, t) + 4t^2 F[e^{\frac{-x^2}{4}}](\omega, t) - 2t^2F[x^2 e^{\frac{-x^2}{4}}](\omega, t)$$
If we then look at a Fourier Transform table we have that,
$$4F[e^{\frac{-x^2}{4}}](\omega, t) + 4t^2 F[e^{\frac{-x^2}{4}}](\omega, t) - 2t^2F[x^2 e^{\frac{-x^2}{4}}](\omega, t) = 4 \sqrt{2}e^{-\omega^2} + 4t^2 \sqrt{2} e^{-\omega^2} - \frac{2 \times 8}{\sqrt{2}} t^3(\frac{1}{2}-\omega^2)e^{-\omega^2} $$
If we do some simplyfication, we arrive at,
$$4 \sqrt{2} e^{-\omega^2}(1+2t^2+\omega^2)$$
So putting this all together, the action of Fourtier Transform on PDE (1) gives,
$$\Omega''(\omega, t)+4\omega^2 \Omega( \omega, t) = 4 \sqrt{2} e^{-\omega^2}(1+2t^2+\omega^2)$$
We now have a linear second order homogeneous equation and this is where I get stuck...
I try to first solve the homogenous part,
$$\Omega''(\omega, t)+4\omega^2 \Omega( \omega, t) = 0$$
But I can't come up with a solution... How does one arrive at a general solution and then a particular and unique solution?