Toronto Math Forum

Verify that the given functions y 1and y 2 satisfy the corresponding homogeneous equation;
then find a particular solution of the given nonhomogeneous equation.
$$
\begin{array}{l}{(1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, 0<t<1} \\ {y_{1}(t)=e^{t}, y_{2}(t)=t}\end{array}
$$

$$
y^{\prime \prime}+\frac{t}{1-t} y^{\prime}-\frac{1}{1-t} y=-2(t-1) e^{-t}
$$
$$
\begin{array}{l}{w=\left|\begin{array}{ll}{e^{t}} & {t} \\ {e^{t}} & {1}\end{array}\right|=e^{t}-t e^{t}} \\ {w_{1}=\left|\begin{array}{ll}{0} & {t} \\ {1} & {1}\end{array}\right|=-t \quad w_{2}=\left|\begin{array}{ll}{e^{t}} & {0} \\ {e^{t}} & {1}\end{array}\right|=e^{\tau}}\end{array}
$$
$$
\begin{aligned} Y(t) &=e^{t} \int \frac{-t \cdot\left(-2(t-1) e^{-t}\right)}{e^{t}-t e^{t}} d t+t \int \frac{e^{t}\left(-2(t-1) e^{-t}\right)}{e^{t}-t e^{t}} d t \\ &=e^{t} \int \frac{-t \cdot\left(t^{2}(t-1) e^{-t}\right)}{e^{t}(1-t)}+t \int \frac{e^{t}\left(t+2(t-1) e^{-t}\right)}{e^{t}(1-t)} d t \\ &=e^{t} \int-2 t e^{-2 t} d t+\int 2 e^{-t} d t \\ &=-2 e^{t} \int t e^{-2 t} d t-2 t e^{-t} \end{aligned}
$$
$$
\begin{array}{ll}{u=t} & {d v=e^{-2 t}} \\ {d u=d t} & {v=-\frac{1}{2} e^{-2 t}}\end{array}
$$
$$
\begin{array}{l}{\displaystyle =-2 e^{t}\left(-\frac{1}{2} e^{-2 t} \cdot t+\int \frac{1}{2} e^{-2 t} d t\right)+t \int 2 e^{-t} d t} \\ {\displaystyle =-2 e^{t}\left(-\frac{1}{2} e^{-2 t} t-\frac{1}{4} e^{-2 t}\right)-2 t e^{-t}} \\ {\displaystyle =t e^{-t}+\frac{1}{2} e^{-t}-2 t e^{-t}} \\ {\displaystyle =\left(\frac{1}{2}-t\right) e^{-t}}\end{array}
$$

$\therefore$ the particular solution of the given non-homogeneous equation is
$$
Y(t)=\left(\frac{1}{2}-t\right) e^{-t}
$$

$
\text{14.}\quad y^{\prime \prime}+4 y^{\prime}+4 y=0, y(-1)=2, y^{\prime}(-1)=1
$

The characteristic equation is,
$$
\begin{aligned} r^{2}+4 r+4 &=0 \\(r+2)^{2} &=0 \\ r_{1}=r_{2} &=-2 \end{aligned}
$$

since it has repeated roots, the general solution is:
$$
\begin{array}{l}{y(t)=c_{1} e^{-2 t}+c_{2} t e^{-2 t}} \\ {y^{\prime}(t)=-2 c_{1} e^{-2 t}+c_{2} e^{-2 t}-2 c_{2} t e^{-2 t}}\end{array}
$$

$\operatorname{plug}$ in $y(-1)=2, \quad y^{\prime}(-1)=1$

we get $$\left\{\begin{array}{l}{c_{1} e^{2}-c_{2} e^{2}=2} \\ {-2 c_{1} e^{2}+c_{2} e^{2}+2 c_{2} e^{2}=1}\end{array}\right.$$

$$
\therefore\left\{\begin{array}{l}{c_{1}=7 e^{-2}} \\ {c_{2}=5 e^{-2}}\end{array}\right.
$$

substitute $C_{2}$ and $C_{2}$ into $y(t)$

$$
=7 e^{-2} e^{-2 t}+5 e^{-2} t e^{-2 t}
$$

$\therefore$ The general solution is $y(t)=7 e^{-2(t+1)}+5 t e^{-2(t+1)}$

7.Find an integrating factor and solve the given equation
$$
\underbrace{1}_{M}+(\underbrace{\frac{x}{y}-\sin(y)}_{N})y^{\prime}=0.
$$


$M_y=0$$N_x=\frac{1}{y}$

$\because$ $M_y\neq N_x$

$\therefore$ not  exact

$R_1=\underbrace{M_y-N_x}_{M'}=\underbrace{0-\frac{1}{y}}_{N'}=-\frac{1}{y}$

$$\left. \begin{array} { l } { \mu = e ^ { - \int R_1 d r } = e ^ { - \int \frac { 1 } { y } } = e ^ { \operatorname { ln } | y | } = y } \\ { \therefore y + ( x - y \operatorname { sin } ( y ) ) y ^ { \prime } = 0 \quad \because M'_y = N'_x = 1 \quad \therefore \text{exact function}} \\{\therefore \exists\quad\varphi(x,y)\qquad s.t \,\varphi_x=M'=y}\\{ \psi = \varphi y d x = y x + h ( y ) }\\{ \varphi y = x + h ^ { \prime } ( y ) = x - y \operatorname { sin } ( y ) }\\{ h ^ { \prime } ( y ) = - y \operatorname { sin } ( y ) }\\{ h ( y ) = - \int y \operatorname { sin } ( y ) }\\{ = y \operatorname { cos } ( y ) - \operatorname { sin } ( y ) }\end{array} \right.$$
$$
\varphi=yx+y\cos(y)-\sin(y)=c
$$