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• Notes

(1) $y {y}^{\prime \prime}=x$

Order $=2$
Degree $=1$

(2) $\left(\frac{d^{4} y}{d x^{4}}\right)^{2}-2\left(\frac{d y}{d x}\right)^{3}=x-\sin x$

Order $=4$
Degree $=2$

(3) $x y^{(7)}-3 y^{\prime \prime}+\sin (x) y^{\prime}-4 y=10$

Order $= 7$
Degree $= 1$

(4) $\frac{1}{x} y^{(5)}-(\ln (x)+x) y^{\prime \prime}-(y-x) y^{\prime}=\frac{\sin x}{x}$

Order $=5$
Degree $=1$

Determine the values of $r$ for which $y=t^{r}, t>0$ is a solution
of the differential equation $t^{2} y^{\prime \prime}-4 t y^{\prime}+6 y=0$

$y=t^{r}$

$y^{\prime}=r t^{r-1}$

$y^{\prime \prime}=r(r-1) t^{r-2}$

$t^{2} y^{\prime \prime}-4 t y^{\prime}+6 y=0$

$t^{2}\left(r(r-1) t^{r-2}\right)-4 t\left(r t^{r-1}\right)+6 y=0$

$t^{2}\left(r^{2}-r\right) t^{r-2}-4 t\left(r t^{r-1}\right)+6 y=0$

$\left(r^{2}-r\right) t^{r}-4 r t^{r}+6 y=0$

$\left(r^{2}-r\right) t^{r}-4 r t^{r}+6 t^{r}=0$

$r^{2}-r-4 r+6=0$

$r^{2}-5 r+6=0$

$\Rightarrow(r-3)(r-2)=0$

${r=3}$

${r=2}$

Assume that $y_{1}(x) \neq 0$ is a solution of $y^{\prime}+p(x) y=q(x)$ for $x>0$
Determine $n$ such that $x y_{1}$ is a solution of $y^{\prime}+\left(p(x)-x^{n}\right) y=x q(x)$

$y_{1}$ is a solution $\longrightarrow y_{1}^\prime+p {y_{1}}=q$

$x y_{1}$ is a solution $\longrightarrow\left(x y_{1}\right)^\prime+\left(p-x^{n}\right)\left(x y_{1}\right)=x q$

$(1)\left(y_{1}\right)+\left(y_{1}^{\prime}\right)(x)+x p y_{1}-x^{n+1} y_{1}=x q$

$y_{1}^{\prime}(x)+x p y_{1}=x q$

$y_{1}^{\prime}+p y_{1}=q$

$y_{1}^{\prime}+p{y_{1}}=q$

$q=q$

$y_{1}-x^{n+1} y_{1}=0$

$\div y_{1}$

$\left[y_{1} \neq 0\right]$

$1-x^{n+1}=0$

$x^{n+1}=1 \quad \Rightarrow x^{n+1}=x^{0}$

$n+1=0 \Rightarrow n=-1$