Author Topic: TUT0801  (Read 636 times)

suyichen

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TUT0801
« on: November 01, 2019, 10:20:25 AM »
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
$$
y(t)=c_{1} e^{t}+c_{2} t e^{t}-\frac{1}{2} \ln \left(1+t^{2}\right) e^{t}+t e^{t} \arctan t
$$