Toronto Math Forum

Find the general solution of the given differential equation.
$$
y^{\prime \prime}-2 y^{\prime}+y=e^{t} /\left(1+t^{2}\right)
$$

$$
\begin{array}{c}{r^{2}-2 r+1=0} \\ {(r-1)(r-1)=0} \\ {r=1,1} \\ {y_{c}=c_{1} e^{t}+c_{2} t e^{t}}\end{array}
$$
$$
\begin{aligned} w &=\left|\begin{array}{cc}{e^{t}} & {t e^{t}} \\ {e^{t}} & {e^{t}+t e^{t}}\end{array}\right| \\ &=e^{2 t}+t e^{2 t}-\left(t e^{2 t}\right) \\ &=e^{2 t} \\
w_{1}&=\left|\begin{array}{cc}{0} & {t e^{t}} \\ {1} & {e^{t}+t e^{t}}\end{array}\right|=-t e^{t}\\
w_{2}&=\left|\begin{array}{cc}{e^{t}} & {0} \\ {e^{t}} & {1}\end{array}\right|=e^{t}
\end{aligned}
$$
$$
Y(t)=e^{t} \int \frac{-t e^{t} \cdot \frac{e^{t}}{1+t^{2}}}{e^{2} t} d t+t e^{t} \int \frac{e^{t} \cdot \frac{e^{t}}{1+t^{2}}}{e^{2} t}
$$
$$
\begin{aligned}-\int \frac{t e^{2 t}}{1+t^{2}} \cdot \frac{1}{e^{2} t} &=-\int \frac{t}{1+t^{2}} d t\qquad\qquad \int \frac{e^{2 t}}{1+t^{2}} \cdot \frac{1}{e^{2 t}}&=\int \frac{1}{1+t^{2}} d t \\ &=-\frac{1}{2} \ln \left(1+t^{2}\right) &=\arctan t\end{aligned}
$$
$$
Y(t)=-\frac{1}{2} e^{t} \ln \left(1+t^{2}\right)+t e^{t} \arctan t
$$

The general solution is
$$
y(t)=c_{1} e^{t}+c_{2} t e^{t}-\frac{1}{2} \ln \left(1+t^{2}\right) e^{t}+t e^{t} \arctan t
$$

Find the solution for $t^{2} y^{\prime \prime}+t y^{\prime}+y=0$

$\displaystyle t^{2} \frac{d y}{d t^{2}}+\alpha t \frac{d y}{d t}+\beta y=0$ is the general form of
Euler equation.

So , we assume $x=\ln t$

$$
\begin{array}{l}{\displaystyle\frac{d y}{d t}=\frac{d y}{d x} \cdot \frac{d x}{d t}=\frac{d y}{d x} \cdot \frac{1}{t}} \\ {\displaystyle\frac{d^{2} y}{d t^{2}}=\frac{1}{t^{2}}\left(\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}\right)}\end{array}
$$

$$
\begin{array}{l}{\therefore t^{2}\left[\frac{1}{t^{2}}\left(\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}\right)\right]+t\left(\frac{1}{t} \frac{d y}{d x}\right)+y=0} \\ {\Rightarrow \frac{d^{2} y}{d x^{2}}+y=0}\end{array}
$$

$$
\therefore \frac{d y}{d x^{2}}+y=0
$$

$$
\begin{array}{l}{\Rightarrow r^{2}+1=0} \\ {r=\sqrt{-1}} \\ {r=\pm 1 i} \\ {x=0 \quad \mu=1}\end{array}
$$

$$
\therefore y(x)=c_{1} e^{0} \cos x+c_{2} e^{0} \sin x
$$

Then, substitute $x=\ln t$.

The solution for the equation is

$$
y(t)=c_{1} \cos (\ln t)+c_{2} \sin (\ln t).
$$

Show that the DE is not exact and become exact mutiplied by the integrating factor. Then find the solution for the DE
$$( \frac { \operatorname { sin } ( y ) } { y } - 2 e ^ { - x } \operatorname { sin } ( x ) ) + ( \frac { \operatorname { cos } ( y ) + 2 e ^ { - x } \operatorname { cos } ( x ) } { y } ) y ^ { \prime } = 0 , \quad \mu ( x , y ) = y e ^ { x };$$

Let $M _ { ( x , y ) } = \frac { \operatorname { sin } y } { y } - 2 e ^ { - x } \operatorname { sin } x \quad N _ { ( x , y ) } = \frac { \operatorname { cos } y + 2 e ^ { - x } \operatorname { cos } x } { y }$

\qquad$M_y = \frac { y \operatorname { cos } y - \operatorname { sin } y } { y ^ { 2 } } \quad N_x = - \frac { 2 } { y } e ^ { - x } \operatorname { cos } x - \frac { 2 } { y } e ^ { - x } \operatorname { sin } x$

Since  $M_y\neq N_x$ , which the given DE is not exact.

So,multipliying thr integrating fact $\mu(x,y)=ye^{x}$
$$\left. \begin{array} { l } { ( e ^ { x } \operatorname { sin } y - 2 y \operatorname { sin } x ) + ( e ^ { x } \operatorname { cos } y + 2 \operatorname { cos } x ) y ^ { \prime } = 0 } \\ { M ^ { \prime } ( x , y ) = e ^ { x } \operatorname { sin } y - 2 y \operatorname { sin } x \quad N ^ { \prime } ( x , y ) = e ^ { x } \operatorname { cos } y + 2 \operatorname { cos } x } \\ { M _ { y } ^ { \prime } = e ^ { x } \operatorname { cos } y - 2 \operatorname { sin } x \quad N ^ { \prime }_x = e ^ { x } \operatorname { cos } y - 2 \operatorname { sin } x } \end{array} \right.$$

Since $M^{\prime}y=N^{\prime}x$, which the DE is exact now.

$$\left. \begin{array}{l}{\exists~\psi _ { ( x , y ) } s.t. \psi_y=N^{\prime}=e^x\cos y+2\cos x}\\{ \psi _ { ( x , y ) } = \int N ^ { \prime } d y = \int e ^ { x } \operatorname { cos } y + 2 \operatorname { cos } x d y }\\{ \qquad~= e ^ { x } \operatorname { sin } y + 2 y \operatorname { cos } x + h ( x ) }\\{ \psi _ { x } = \operatorname { sin } y e ^ { x } - 2 y \operatorname { sin } x \cdot \operatorname { th } ^ { \prime } ( x ) = M ^ { \prime } }\\{\therefore h^{\prime}(x)=0\quad h(x)=c}\\{\therefore \psi(x,y)=e^x\sin y+2y\cos x}\end{array} \right.$$

Therefore, the solution of the DE is
$$
e^x\sin y+2y\cos y=c
$$