I'm wondering if somebody can help me with this question.
Construct a second-order linear differential equation that has this following general solution:
$$x(t)=a(t^2-1)+be^{t^2+t}$$ for a,b two constant.
I tried to compute : $\frac{d^2}{dt^2}x(t)$ and $\frac{d}{dt}x(t)$ and after put it in this following equation:
$$\frac{d^2}{dt^2}x(t)+ a_1(t)\frac{d}{dt}x(t) + a_2(t)x(t) = u(t)$$
And then find $a_1(t),a_2(t)$ and $u(t)$ such that is it true. Hence we obtain :
$$a(2+a_1(t)2t + a_2(t)(t^2-1))+b(2e^{t^2+t} + (2t +1)e^{t^2+t} + a_1(t)(2t+1)e^{t^2+t} + a_2(t)e^{t^2+t}) = u(t)$$ Since I don't want a particular equation but one which is true, I consider that the parenthesis of b is equal to zero, and so I found :
$$\frac{d^2}{dt^2}x(t) - (2t + 1)\frac{d}{dt}x(t) -2 x(t) = -a(6t^2-2t-3).$$
But I think it's not true, because there is always our constant a.
Given two linearly independent solutions $\phi_1$ and $\phi_2$ there is always a corresponding homogenous differential equation $$ W(\phi_1,\phi_2, x) = W(\phi_1,\phi_2) x'' - W'(\phi_1,\phi_2) x' + W(\phi_1',\phi_2') x =0, $$ where $W$ denotes the Wronski determinant. More information can be found in Section 3.5 on my book on Ordinary Differential Equations.