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Consider the equation $$ty''-(t-1)y'+y=t^2e^{2t} $$ (eq.1)

Knowing that $$ y_1(t) = e^t $$ is a general solution of the corresponding homogeneous differential equation, determine the general solution of the homogeneous equation.

Checking the solution: $$ te^t-(t-1)e^t+e^t=0$$ $$ 0=0 $$

Making $$ y=uy_1 = ue^t$$ $$y'=e^tu+e^tu' $$ $$ y''= e^tu+u'e^t+u''e^t+u'e^t$$

Substituting in equation 1

$$u''te^t+u'(te^t+e^t)+2ue^t=0$$ Putting $$v=u'$$ $$v'(te^t)+v(te^t+e^t)+2u=0$$

simplifying $$v't+v(t+1)+2u=0$$

Now I don't know how to continue. Can you help? Thanks

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    $\begingroup$ there is a sign mistake somewhere in your equation because $e^t$ is not a solution to the homogeneous equation. Check it $$te^t-(t-1)e^t+e^t=0$$ it gives $2e^t=0$ $\endgroup$ – Isham Dec 1 at 16:12
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    $\begingroup$ Yes, the equation inicial is $$ ty′′−(t+1)y′+y=t^2e^{2t} $$. It was my mistake $\endgroup$ – Marta Seara Dec 1 at 16:36
  • $\begingroup$ Then I answered the question. $\endgroup$ – Isham Dec 1 at 16:36
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Assuming the DE as

$$ t y''-(t+1)y'+y=t^2e^{2t} $$

here $y_1 = e^t$ is a solution for the homogeneous DE. It is easy to find also a polynomial solution for the homogeneous DE with $y_2 = t+1$ so

$$ y_h = c_1 (t+1)+ c_2 e^t $$

now making $y_p = c_1(t)(t+1)+c_2(t)e^t$ after substitution in the complete DE we arrive at

$$ -(t^2+1)c_1'+t(t+1)c_1'' +e^t(t-1)c_2'+t e^2c_2''-t^2e^{2t}=0 $$

here $c_1(t),c_2(t)$ are independent functions so we choose

$$ \cases{-(t^2+1)c_1'+t(t+1)c_1''= 0\\ e^t(t-1)c_2'+t e^tc_2''- t^2e^{2t}=0} $$

or

$$ \cases{-(t^2+1)c_1'+t(t+1)c_1''= 0\\ (t-1)c_2'+t c_2''=t^2e^t} $$

and now making $u=c_1', v= c_2'$ we have

$$ \cases{-(t^2+1)u+t(t+1)u'= 0\\ (t-1)v+t v'=t^2e^t} $$

and then choosing $u=0$, $v = \frac 12 t e^t$ we have, (we are choosing a particular solution so we select the convenient value for the integration constants)

$$ \cases{c_1(t) = 0\\ c_2(t) = \frac 12(t-1)e^t } $$

and finally

$$ y = y_h+y_p = c_1 (t+1)+ c_2 e^t + 0(t+1)+\frac 12(t-1)e^t e^t $$

or

$$ y = c_1 (t+1)+ c_2 e^t +\frac 12(t-1)e^{2t} $$

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  • $\begingroup$ Thanks for the help. Just one more small question, I understand that Y2=t+1 is another homogeneous solution, but how can I prove it? Can you help me? $\endgroup$ – Marta Seara Dec 2 at 21:33
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    $\begingroup$ A second order homogeneous DE admits two independent solutions. We know already $y_1=e^t$. By pure inspection, making $y_2 = a t + b$ and substituting into the homogeneous we have $0-(t+1)a+at+b=0$ so we obtain the condition $a-b=0$ so we choose $a=b=1$ giving $y_2 = t+1$ This function is independent of $y_1= e^t$ so it is a feasible independent solution for the homogeneous also. $\endgroup$ – Cesareo Dec 2 at 22:23
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$$ty''-(t-1)y'+y=t^2e^{2t}$$ $e^t$ is not a solution of the homogeneous solution that's why trying to reduce the order of the equation didn't work.Maybe you made a sign error and the equation is : $$ty''-(t+1)y'+y=t^2e^{2t}$$ Here you can try reduction of order method. With $y=ve^t$. $$\implies tv''+v'(t-1)=t^2e^t$$ Susbtitute $u=v'$ and solve as a first order DE. Or write it as: $$\left (\frac {v'}t\right )'+\frac {v'}t=e^t$$ $$\left (e^t\frac {v'}t\right )'=e^{2t}$$ Integrate and substitute back $y=ve^t$

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  • $\begingroup$ And if I substitute for u=v′ and solve as a first order DE $\endgroup$ – Marta Seara Dec 1 at 17:05
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    $\begingroup$ It's great too. Marta. There are many methods for solving this kind of equation. Choose the one that you prefer. $\endgroup$ – Isham Dec 1 at 17:07
  • $\begingroup$ Thanks for the help. Just one more small question, I understand that Y2=t+1 is another homogeneous solution, but how can I prove it? Can you help me? $\endgroup$ – Marta Seara Dec 2 at 22:09
  • $\begingroup$ You can plug the solution $y_2$ in the Differential equation to prove its a solution @MartaSeara It's a solution of the homogeneous equation $\endgroup$ – Isham Dec 2 at 22:11

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