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I have this problem involving second order linear differential equations. I know how to solve a first order one, but I've been searching for an answer for this one and I don't seem to find any complete solution. Here's the equation: $$t^2\cdot x''- t\cdot x' +4\cdot x= \log(t),\;t>0$$ Can someone help, please?

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  • $\begingroup$ y=y(t),y=y(x) or y=y(x,t)? $\endgroup$ – johny Jun 3 '18 at 11:04
  • $\begingroup$ Sorry, I got it mixed up. I edited it. $\endgroup$ – user372244 Jun 3 '18 at 11:07
  • $\begingroup$ x=x(t) of course $\endgroup$ – user372244 Jun 3 '18 at 11:08
  • $\begingroup$ Hi and welcome to the site, I helped with some typesetting, there is a tutorial on mathjax typesetting somewhere on here, or you can click edit on your question to see how it is done. here is tutorial math.meta.stackexchange.com/questions/5020/… $\endgroup$ – mathreadler Jun 3 '18 at 11:09
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This can be solved using a substitution by writing $t=e^u$. $$t=e^u\to\frac{dt}{du}=e^u=t$$ $$\frac{dx}{du}=\frac{dt}{du}\frac{dx}{dt}=t\frac{dx}{dt}$$ Then it can be written that $$\begin{equation}\begin{aligned} \frac{d^2x}{du^2}=\frac{d}{du}\bigg[\frac{dx}{du}\bigg]&=\frac{d}{dt}\bigg[t\frac{dx}{dt}\bigg]\times\frac{dt}{du} \\ &=\bigg(t\frac{d^2x}{dt^2}+\frac{dx}{dt}\bigg)\times\frac{dt}{du} \\ &=t^2\frac{d^2x}{dt^2}+t\frac{dx}{dt} \end{aligned}\end{equation}$$ And so the equation can be rewritten to the form $$\frac{d^2x}{du^2}-2t\frac{dx}{dt}+4x=\log e^u$$ $$\frac{d^2x}{du^2}-2\frac{dx}{du}+4x=u$$ From this it can be solved as a linear inhomogenous second order differential equation.

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$$t^2\cdot x''- t\cdot x' +4\cdot x= \ln(t), t>0$$ Substitute $x=t^m$ to solve the homogeneous equation $$m^2-2m+4=0 \implies m=1 \pm i\sqrt 3$$ $$ \implies x(t)=C_1t\cos (\sqrt 3 \ln t)+C_2t\sin (\sqrt 3 \ln t)$$

Then use variation of constants for the inhomogeneous equation

$$x=A\ln(t)+B$$

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