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Term Test 2 / Re: Problem 1 (main sitting)
« on: November 19, 2019, 05:24:49 AM »
Solution see attachment.
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Quiz-5 / TUT0202 Quiz5
« on: November 01, 2019, 02:00:08 PM »
Verify that the given functions y 1and y2 satisfy the corresponding homogeneous equation, then find a particular solution of the given nonhomogeneous equation.
$$
t^{2} y^{\prime \prime}-t(t+2) y^{\prime}+(t+2) y=2 t^{3}, t>0 ; y_{1}(t)=t, y_{2}(t)=t e^{t}
$$
$$
y_{1}(t)=t \quad y_{1}^{\prime}(t)=1 \quad y_{1}^{\prime \prime}(t)=0
$$
$$
-t(t+2)+(t+2) t=0
$$
$$
y_{2}(t)=t e^{t} \quad y_{2}^{\prime}(t)=e^{t}+t e^{t} \quad y_{2}^{\prime \prime}(t)=2 e^{t}+t e^{t}
$$
$$
t^{2}\left(2 e^{t}+t e^{t}\right)-t(t+2)\left(e^{t}+t e^{t}\right)+(t+2) t e^{t}=0
$$
Hence, $y_{1}$ and $y_{2}$ satisfy the homogeneons equation
$$
y^{\prime \prime}-\frac{t+2}{t} y^{\prime}+\frac{t+2}{t^{2}} y=2 t
$$
$$
\begin{aligned}
w&=\left|\begin{array}{cc}{t} & {t e^{t}} \\ {1} & {e^{t}+\operatorname{te}^{t}}\end{array}\right|=t^{2} e^{t}\\
w_{1}&=\left|\begin{array}{cc}{0} & {t e^{t}} \\ {1} & {e^{t}+t e^{t}}\end{array}\right|=-t e^{t}\\
w_{2}&=\left|\begin{array}{cc}{t} & {0} \\ {1} & {1}\end{array}\right|=t
\end{aligned}
$$
$$
\begin{aligned} y_{p}(t) &=t \int \frac{\left(-t e^{t}\right)(2 t)}{t^{2} e^{t}} d t+t e^{t} \int \frac{(t)(2 t)}{t^{2} e^{t}} d t \\ &=t \int-2 d t+2 t e^{t} \int \frac{1}{e^{t}} d t \\ &=t(-2 t)+2 t e^{t}\left(-e^{-t}\right) \\ &=-2 t^{2}-2 t \end{aligned}
$$
$$
t^{2} y^{\prime \prime}-t(t+2) y^{\prime}+(t+2) y=2 t^{3}, t>0 ; y_{1}(t)=t, y_{2}(t)=t e^{t}
$$
$$
y_{1}(t)=t \quad y_{1}^{\prime}(t)=1 \quad y_{1}^{\prime \prime}(t)=0
$$
$$
-t(t+2)+(t+2) t=0
$$
$$
y_{2}(t)=t e^{t} \quad y_{2}^{\prime}(t)=e^{t}+t e^{t} \quad y_{2}^{\prime \prime}(t)=2 e^{t}+t e^{t}
$$
$$
t^{2}\left(2 e^{t}+t e^{t}\right)-t(t+2)\left(e^{t}+t e^{t}\right)+(t+2) t e^{t}=0
$$
Hence, $y_{1}$ and $y_{2}$ satisfy the homogeneons equation
$$
y^{\prime \prime}-\frac{t+2}{t} y^{\prime}+\frac{t+2}{t^{2}} y=2 t
$$
$$
\begin{aligned}
w&=\left|\begin{array}{cc}{t} & {t e^{t}} \\ {1} & {e^{t}+\operatorname{te}^{t}}\end{array}\right|=t^{2} e^{t}\\
w_{1}&=\left|\begin{array}{cc}{0} & {t e^{t}} \\ {1} & {e^{t}+t e^{t}}\end{array}\right|=-t e^{t}\\
w_{2}&=\left|\begin{array}{cc}{t} & {0} \\ {1} & {1}\end{array}\right|=t
\end{aligned}
$$
$$
\begin{aligned} y_{p}(t) &=t \int \frac{\left(-t e^{t}\right)(2 t)}{t^{2} e^{t}} d t+t e^{t} \int \frac{(t)(2 t)}{t^{2} e^{t}} d t \\ &=t \int-2 d t+2 t e^{t} \int \frac{1}{e^{t}} d t \\ &=t(-2 t)+2 t e^{t}\left(-e^{-t}\right) \\ &=-2 t^{2}-2 t \end{aligned}
$$
3
Quiz-5 / TUT0202 Quiz5
« on: November 01, 2019, 09:57:46 AM »
Verify that the given functions y 1and y2 satisfy the corresponding homogeneous equation;
then find a particular solution of the given nonhomogeneous equation.
$$
t^{2} y^{\prime \prime}-t(t+2) y^{\prime}+(t+2) y=2 t^{3}, t>0 ; y_{1}(t)=t, y_{2}(t)=t e^{t}
$$
$$
y_{1}(t)=t \quad y_{1}^{\prime}(t)=1 \quad y_{1}^{\prime \prime}(t)=0
$$
$$
-t(t+2)+(t+2) t=0
$$
$$
y_{2}(t)=t e^{t} \quad y_{2}^{\prime}(t)=e^{t}+t e^{t} \quad y_{2}^{\prime \prime}(t)=2 e^{t}+t e^{t}
$$
$$
t^{2}\left(2 e^{t}+t e^{t}\right)-t(t+2)\left(e^{t}+t e^{t}\right)+(t+2) t e^{t}=0
$$
Hence, $y_{1}$ and $y_{2}$ satisfy the homogeneons equation
$$
y^{\prime \prime}-\frac{t+2}{t} y^{\prime}+\frac{t+2}{t^{2}} y=2 t
$$
$$
\begin{aligned}
w&=\left|\begin{array}{cc}{t} & {t e^{t}} \\ {1} & {e^{t}+\operatorname{te}^{t}}\end{array}\right|=t^{2} e^{t}\\
w_{1}&=\left|\begin{array}{cc}{0} & {t e^{t}} \\ {1} & {e^{t}+t e^{t}}\end{array}\right|=-t e^{t}\\
w_{2}&=\left|\begin{array}{cc}{t} & {0} \\ {1} & {1}\end{array}\right|=t
\end{aligned}
$$
$$
\begin{aligned} y_{p}(t) &=t \int \frac{\left(-t e^{t}\right)(2 t)}{t^{2} e^{t}} d t+t e^{t} \int \frac{(t)(2 t)}{t^{2} e^{t}} d t \\ &=t \int-2 d t+2 t e^{t} \int \frac{1}{e^{t}} d t \\ &=t(-2 t)+2 t e^{t}\left(-e^{-t}\right) \\ &=-2 t^{2}-2 t \end{aligned}
$$
then find a particular solution of the given nonhomogeneous equation.
$$
t^{2} y^{\prime \prime}-t(t+2) y^{\prime}+(t+2) y=2 t^{3}, t>0 ; y_{1}(t)=t, y_{2}(t)=t e^{t}
$$
$$
y_{1}(t)=t \quad y_{1}^{\prime}(t)=1 \quad y_{1}^{\prime \prime}(t)=0
$$
$$
-t(t+2)+(t+2) t=0
$$
$$
y_{2}(t)=t e^{t} \quad y_{2}^{\prime}(t)=e^{t}+t e^{t} \quad y_{2}^{\prime \prime}(t)=2 e^{t}+t e^{t}
$$
$$
t^{2}\left(2 e^{t}+t e^{t}\right)-t(t+2)\left(e^{t}+t e^{t}\right)+(t+2) t e^{t}=0
$$
Hence, $y_{1}$ and $y_{2}$ satisfy the homogeneons equation
$$
y^{\prime \prime}-\frac{t+2}{t} y^{\prime}+\frac{t+2}{t^{2}} y=2 t
$$
$$
\begin{aligned}
w&=\left|\begin{array}{cc}{t} & {t e^{t}} \\ {1} & {e^{t}+\operatorname{te}^{t}}\end{array}\right|=t^{2} e^{t}\\
w_{1}&=\left|\begin{array}{cc}{0} & {t e^{t}} \\ {1} & {e^{t}+t e^{t}}\end{array}\right|=-t e^{t}\\
w_{2}&=\left|\begin{array}{cc}{t} & {0} \\ {1} & {1}\end{array}\right|=t
\end{aligned}
$$
$$
\begin{aligned} y_{p}(t) &=t \int \frac{\left(-t e^{t}\right)(2 t)}{t^{2} e^{t}} d t+t e^{t} \int \frac{(t)(2 t)}{t^{2} e^{t}} d t \\ &=t \int-2 d t+2 t e^{t} \int \frac{1}{e^{t}} d t \\ &=t(-2 t)+2 t e^{t}\left(-e^{-t}\right) \\ &=-2 t^{2}-2 t \end{aligned}
$$
4
Term Test 1 / Re: Problem 3 (main)
« on: October 23, 2019, 08:04:31 AM »
Find the general solution of the equation
$$
y^{\prime \prime}-2 y^{\prime}-3 y=16 \cosh x
$$
$$
\begin{aligned} r^{2}-2 r-3 &=0 \\(r-3)(r+1) &=0 \\ r=3, r&=-1 \end{aligned}
$$
$$
\begin{array}{l}{y_{c}=c_{1} e^{3 x}+c_{2} e^{-x}} \\ {y^{\prime \prime}-2 y^{\prime}-3 y=16 \cosh x=8 e^{x}+8 e^{-x}} \\ {y^{\prime \prime}-2 y^{\prime}-3 y=8 e^{x}}\end{array}
$$
Let $y_{p}=A e^{x}\quad y^{\prime}=A e^{x} \quad y^{\prime \prime}=A e^{x}$
$$\begin{aligned} A e^{x}-2 A e^{x}-3 A e^{x} &=8 e^{x} \\-4 A e^{x} &=8 e^{x} \\ A &=-2 \end{aligned}$$
$$
\therefore y_{p}=-2 e^{x}
$$
$$
y^{\prime \prime}-2 y^{\prime}-3 y=8 e^{-x}
$$
$\operatorname{let} y_{p}(t)=A x e^{-x} \quad y^{\prime}=A e^{-x}-A x e^{-x}$
$$
\begin{aligned} y^{\prime \prime}=&-A e^{-x}-\left(A e^{-x}-A x e^{-x}\right) \\ &=-2 A e^{-x}+A x e^{-x} \end{aligned}
$$
$$
-2 A e^{-x}+A x e^{-x}-2 A e^{-x}+2 A x e^{-x}-3 A x e^{-x}=8 e^{-x}
$$
$$
\begin{aligned}-2 A-2 A &=8 \\ A &=-2 \\ y_{p} &=-2 \times e^{-x} \end{aligned}
$$
$$
y=c_{1} e^{3 x}+c_{2} e^{-x}-2 e^{x}-2 x e^{-x}
$$
$$
y^{\prime \prime}-2 y^{\prime}-3 y=16 \cosh x
$$
$$
\begin{aligned} r^{2}-2 r-3 &=0 \\(r-3)(r+1) &=0 \\ r=3, r&=-1 \end{aligned}
$$
$$
\begin{array}{l}{y_{c}=c_{1} e^{3 x}+c_{2} e^{-x}} \\ {y^{\prime \prime}-2 y^{\prime}-3 y=16 \cosh x=8 e^{x}+8 e^{-x}} \\ {y^{\prime \prime}-2 y^{\prime}-3 y=8 e^{x}}\end{array}
$$
Let $y_{p}=A e^{x}\quad y^{\prime}=A e^{x} \quad y^{\prime \prime}=A e^{x}$
$$\begin{aligned} A e^{x}-2 A e^{x}-3 A e^{x} &=8 e^{x} \\-4 A e^{x} &=8 e^{x} \\ A &=-2 \end{aligned}$$
$$
\therefore y_{p}=-2 e^{x}
$$
$$
y^{\prime \prime}-2 y^{\prime}-3 y=8 e^{-x}
$$
$\operatorname{let} y_{p}(t)=A x e^{-x} \quad y^{\prime}=A e^{-x}-A x e^{-x}$
$$
\begin{aligned} y^{\prime \prime}=&-A e^{-x}-\left(A e^{-x}-A x e^{-x}\right) \\ &=-2 A e^{-x}+A x e^{-x} \end{aligned}
$$
$$
-2 A e^{-x}+A x e^{-x}-2 A e^{-x}+2 A x e^{-x}-3 A x e^{-x}=8 e^{-x}
$$
$$
\begin{aligned}-2 A-2 A &=8 \\ A &=-2 \\ y_{p} &=-2 \times e^{-x} \end{aligned}
$$
$$
y=c_{1} e^{3 x}+c_{2} e^{-x}-2 e^{x}-2 x e^{-x}
$$
5
Quiz-3 / TUT0202 Quiz3
« on: October 11, 2019, 02:00:30 PM »
Find a differential equation whose general solution is $y=c_{1} e^{-t / 2}+c_{2} e^{-2 t}$.
Then $r_{1}=-\frac{1}{2}, r_{2}=-2$ are two roots of the characteristic equation of the required differential equation.
so the characteristic equation should look like
$$
\begin{aligned}\left(r+\frac{1}{2}\right)(r+2) &=0 \\(2 r+1)(r+2) &=0 \\ 2 r^{2}+5 r+2 &=0 \end{aligned}
$$
which corresponds to the DE
$$
2 y^{\prime \prime}+5 y^{\prime}+2 y=0
$$
Then $r_{1}=-\frac{1}{2}, r_{2}=-2$ are two roots of the characteristic equation of the required differential equation.
so the characteristic equation should look like
$$
\begin{aligned}\left(r+\frac{1}{2}\right)(r+2) &=0 \\(2 r+1)(r+2) &=0 \\ 2 r^{2}+5 r+2 &=0 \end{aligned}
$$
which corresponds to the DE
$$
2 y^{\prime \prime}+5 y^{\prime}+2 y=0
$$
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