Find a solution to the differential equation: Which is NOT a constant multiple of the solution given.

Question

Find a solution to the equation: $$ ty'' - (t+1)y' + y =0 $$ Which is not a constant multiple of the solution: $y(t) = e^t$.

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Hi, I have worked this problem and have solved it, using the particular solution of $e^t$. Why is this problem asking for a solution that is not a constant multiple of the only given solution? I do not see another way of solving it

Thank you!

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(solution from comment)

Using reduction of order we aim for finding $y_2$ based on the solution given. In this case $y(t) = e^t$ is a solution. Following that we have $y_2 = g(t) y_1$. $G$ is an unknown and will eventually be substituted for $u = g'$. Following this we differentiate $y_2$ to get $$y_2' = g_1' e^t + g_1 e^t.$$ We differentiate again for $$y_2'' = g_1'' e^t + 2g_2' e^t + g_2 e^t.$$ Next, we substitute $y_2$, $y_2'$ and $y_2''$ into the given equation and also substitute $u=g'$ to get $$tu' +2ut -tu -u =0.$$ Solve this equation by separation and then integrate and get the solution of y_2. We find the last $y_2 = t+1$.

Answer 1

Write $$ t(y''-y') = y'-y$$

Now let $z=y'-y$, then $${z'\over z} = {1\over t}$$

so $$ \ln z = \ln t+ c\implies z = a\cdot t$$

for some real $a$. Now you can try if $y$ can be linear $y=kt+n$:

$$ k-kt-n = at \;\;\; \implies \;\;\; k=-a\;\;\wedge \;\;n=k=-a$$

so $y= -at-a$ is solution for every real $a$.

Answer 2

Well, one way we could go is by power series. Letting $y=\sum_{k=0}^\infty a_kt^k,$ we have: $$\begin{eqnarray}ty''&=&t\sum_{k=2}^\infty k(k-1)a_kt^{k-2}\\&=&\sum_{k=1}^\infty(k+1)ka_{k+1}t^k\end{eqnarray}\tag{1}$$ $$\begin{eqnarray}(t+1)y'&=&(t+1)\sum_{k=1}^\infty ka_kt^{k-1}\\&=&\sum_{k=1}^\infty ka_kt^k+\sum_{k=1}^\infty ka_kt^{k-1}\\&=&a_1+\sum_{k=1}^\infty ka_kt^k+\sum_{k=2}^\infty ka_kt^{k-1}\\&=&a_1+\sum_{k=1}^\infty \left(ka_k+(k+1)a_{k+1}\right)t^k\end{eqnarray}\tag{2}$$

Putting this into the given equation, we have $$0=a_0-a_1+\sum_{k=1}^\infty\left((k+1)ka_{k+1}-ka_k-(k+1)a_{k+1}+a_k\right)t^k,$$ or $$0=a_0-a_1+\sum_{k=2}^\infty(k-1)\left((k+1)a_{k+1}-a_k\right)t^k.$$ Consequently, we must have $a_0-a_1=0,$ and for all $k\ge 2,$ we must have $$(k+1)a_{k+1}-a_k=0.$$ Put another way, we must have $a_1=a_0,$ and for $k\ge 2,$ we must have $a_{k+1}=\frac1{k+1}a_k.$ Thus, all of the coefficients are completely determined by the values of $a_0$ and $a_2,$ since by an easy inductive argument, we have for all $k\ge2$ that $$a_k=\frac2{k!}a_2.$$

Thus, we have $$y=\sum_{k=0}^\infty a_kt^k=a_0(1+t)+2a_2\sum_{k=2}^\infty\frac1{k!}t^k,$$ and so $$y=a_0(1+t)+2a_2\left(e^t-1-t\right)=(a_0-2a_2)(1+t)+2a_2e^t=c_1(1+t)+c_2e^t.$$ Thus, a particular solution that is not a constant multiple of $e^t$ is (for example) $t+1.$

Answer 3

$$ty'' - (t+1)y' + y =0$$ $$ty'' - ty'-1y' + y =0$$ $$ty'' -y'-( ty'-y) =0$$ Divide by the integrating factor $\mu(t)=t^2$ $$\frac {ty'' -y'}{t^2}-\frac {( ty'-y)}{t^2} =0$$

$$(\frac {y'}t)'-(\frac yt)' =0$$ Integrate

$$\frac {y'}t-\frac yt =K_1$$ $$(ye^{-t})'=K_1te^{-t}$$ Integrate again $$ye^{-t}=-K_1(t+1)e^{-t}+K_2$$ $$\implies y(t)=C_1(t+1)+C_2e^t$$