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Find the general solution for $t > 0$ using variation of parameters (look for fundamental set of homogeneous solutions $t^r$

$$t^2 x'' + 7tx' + 5x = t$$

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  • $\begingroup$ Have you started doing this task on your own? $\endgroup$ – svavil Nov 7 '15 at 22:00
  • $\begingroup$ I tried solving the homogeneous equation but I don't know how to solve it with non-constant coefficients $\endgroup$ – user286826 Nov 7 '15 at 22:20
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For the homogeneous solutions, you need to solve

$$t^2x''+7tx'+5x=0$$

Put $x=t^r$ with $r$ a fixed constant to be determined, then $x''=r(r-1)x^{r-2}$ and $x'=rx^{r-1}$.

Now you should b able to substitute into the homogeneous D.E. to get an equation involving $x^r$ on the left-hand side and zero on the right-hand side. Hence get an equation for the coefficient of $x^r$ and get a quadratic equation for $r$.


Note that the original D.E. has a particular solution $x_p=\frac{1}{12}t$

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  • $\begingroup$ If $t=x^r$ then $x''=r(r-1)x^{r-2}$ and $x'=rx^{r-1}$ and what is $x$? I would like to better understand how to solve such DFE's. $\endgroup$ – zoli Nov 7 '15 at 23:24
  • $\begingroup$ @zoli - $x=x(t)$ is simply a function of $t$. I can see why you were confused - for some reason I wrote $t=x^r$ when meaning $x=t^r$ (as per the question). This is the hint in the question, and arises because the form of the homogeneous D.E. suggests a polynomial solution, i.e. for every successive derivative, the coefficient is one degree higher. $\endgroup$ – Marconius Nov 8 '15 at 0:45
  • $\begingroup$ So edited you. Thank you. $\endgroup$ – zoli Nov 8 '15 at 0:47

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