If $x=(a+bt)e^{-nt}$, show that $\frac{d^2x}{dt^2} + 2n\frac{dx}{dt} +n^2x = 0$

Question

Question : If $x=(a+bt)e^{-nt}$, show that $\frac{d^2x}{dt^2} + 2n\frac{dx}{dt} +n^2x = 0$

I got $\frac{dx}{dt}$
Then $\frac{d^2x}{dt^2}$ and plugged in these values

But at the end I got $$-n^2ae^{-nt} -n^2bte^{-nt} +n^2x$$

I don't know how to proceed further

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How I proceeded with this

I simplified the bracket further $(x=ae^{-nt} + bte^{-nt})$

And solved for $\frac {dx} {dt}$ ( i got $\frac {dx} {dt}$ = $-nae^{-nt} - nbte^{-nt} + be^{-nt}$

And then solved for $\frac {d^2x} {dt^2}$

( i got $\frac {d^2x} {dt^2}$ = $n^2ae^{-nt} + n^2bte^-nt - nbe^{-nt} - nbe^{-nt})$

Answer 1

$$xe^{nt}=at+b$$

Method$\#1:$

Differentiate wrt $x$ $$e^{nt}\left(nx+\dfrac{dx}{dt}\right)=a$$

Differentiate wrt $x$ $$e^{nt}n\left(nx+\dfrac{dx}{dt}\right)+e^{nt}\left(n\dfrac{dx}{dt}+\dfrac{d^2x}{dt^2}\right)=0$$

Now safely cancel $e^{nt}\ne0$

Method$\#2:$

General Leibniz rule with $n=2$

$$\dfrac{d^2x}{dt^2}e^{nt}+\binom21\dfrac{dx}{dt}\dfrac{d(e^{nt})}{dt}+\binom22x\dfrac{d^2(e^{nt})}{dt^2}=0$$

Answer 2

You can also show that $$\exp(-nt)\,\frac{\text{d}^2}{\text{d}t^2}\,\Big(\exp(nt)\,f(t)\Big)=\frac{\text{d}^2}{\text{d}t^2}\,f(t)+2n\,\frac{\text{d}}{\text{d}t}\,f(t)+n^2\,f(t)$$ for all twice differentiable function $f$. One way to see this is to write $$\exp(-nt)\,\frac{\text{d}}{\text{d}t}\,\Big(\exp(nt)\,g(t)\Big)=\frac{\text{d}}{\text{d}t}\,g(t)+n\,g(t)\,,$$ and $$\exp(-nt)\,\frac{\text{d}^2}{\text{d}t^2}\,\Big(\exp(nt)\,f(t)\Big)=\exp(-nt)\,\frac{\text{d}}{\text{d}t}\,\exp(nt)\,\Biggl(\exp(-nt)\,\frac{\text{d}}{\text{d}t}\,\Big(\exp(nt)\,f(t)\Big)\Biggr)\,.$$

Thus, $$\frac{\text{d}^2}{\text{d}t^2}\,f(t)+2n\,\frac{\text{d}}{\text{d}t}\,f(t)+n^2\,f(t)=0$$ if and only if $\exp(nt)\,f(t)=a+bt$ for some constants $a$ and $b$, or equivalently, $$f(t)=(a+bt)\,\exp(-nt)\,.$$