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Term Test 2 / Re: Problem 4 (noon)
« on: November 19, 2019, 09:21:22 AM »
Please see the attachment for the answer
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2
Term Test 2 / Re: Problem 3 (noon)
« on: November 19, 2019, 09:15:02 AM »
Please see the attachment for the answer
3
Term Test 2 / Re: Problem 2 (noon)
« on: November 19, 2019, 09:11:36 AM »
Please see the attachment for the answer
4
Term Test 2 / Re: Problem 1 (noon)
« on: November 19, 2019, 09:04:42 AM »
Please see the attachment for the answer
OK.
But No snapshots!
OK.
But No snapshots!
5
Quiz-5 / TUT 0401 Quiz 5
« on: November 01, 2019, 03:46:25 PM »
\begin{equation}
\begin{array}{c}{\text { Qusetion: }} \\ {\qquad \begin{array}{c}{(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}}\end{array}
\end{equation}
\begin{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} \\ {\text { Hence, }} \\ {\qquad\left\{\begin{array}{l}{y_{1}(t)=e^{t}} \\ {y_{1}^{\prime}(t)=e^{t}} \\ {y_{1}^{\prime \prime}(t)=e^{t}}\end{array} \text { and }\left\{\begin{array}{l}{y_{2}(t)=t} \\ {y_{2}^{\prime \prime}(t)=1} \\ {y_{2}^{\prime \prime}(t)=0}\end{array}\right.\right.}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Substitute back into the homogeneous equation: }} \\ {\qquad(1-t) y^{\prime \prime}+t y^{\prime}-y=0} \\ {\text { Verfified that } y_{1}(t) \text { and } y_{2}(t) \text { both satisfy the corresponding homogeneous }} \\ {\text { equation. }}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { And the complementary solution } y_{c}(t)=c_{1} e^{t}+c_{2}} \\ {\text { Now divide both sides of the original equation by } 1-t:} \\ {\qquad y^{\prime \prime}+\frac{t}{1-t} y'-\frac{1}{1-t} y=-2(t-1) e^{-t}} \\ {\qquad p(t)=\frac{t}{1-t}, g(t)=-2(t-1) e^{-t}} \\ {\qquad W\left[y_{1}, y_{2}\right](t)=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right|=(1-t) e^{t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { since the particular solution has the form: }} \\ {\qquad Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)}\end{array}
\end{equation}
\begin{equation}
\begin{aligned} u_{1}(t) &=-\int \frac{y_{2}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t \\ &=-\int \frac{t \cdot\left(-2(t-1) e^{-t}\right)}{(1-t) e^{t}} d t \\ &=-2 \int t e^{-2 t} d t \\ &=\left(t+\frac{1}{2}\right) e^{-2 t} \end{aligned}
\end{equation}
\begin{equation}
\begin{aligned} u_{2}(t) &=\int \frac{y_{1}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t \\ &=\int \frac{e^{t^{t} \cdot\left(-2(t-1) e^{-t}\right)}}{(1-t) e^{t}} d t \\ &=2 \int e^{-t} \\ &=-2 e^{-t} \end{aligned}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Therefore, }} \\ {\qquad Y(t)=\left(t+\frac{1}{2}\right) e^{-2 t} \cdot e^{t}+\left(-2 e^{-t}\right) \cdot t=\left(\frac{1}{2}-t\right) e^{-t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Hence, the general solution: }} \\ {\qquad \begin{aligned} y(t) &=y_{c}(t)+Y(t) \\ &=c_{1} e^{t}+c_{2} t+\left(\frac{1}{2}-t\right) e^{-t} \end{aligned}} \\ {\text { Therefore, the particular solution of the given nonhomogeneous equation is }} \\ {\qquad Y(t)=\left(\frac{1}{2}-t\right) e^{-t}}\end{array}
\end{equation}
\begin{array}{c}{\text { Qusetion: }} \\ {\qquad \begin{array}{c}{(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}}\end{array}
\end{equation}
\begin{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} \\ {\text { Hence, }} \\ {\qquad\left\{\begin{array}{l}{y_{1}(t)=e^{t}} \\ {y_{1}^{\prime}(t)=e^{t}} \\ {y_{1}^{\prime \prime}(t)=e^{t}}\end{array} \text { and }\left\{\begin{array}{l}{y_{2}(t)=t} \\ {y_{2}^{\prime \prime}(t)=1} \\ {y_{2}^{\prime \prime}(t)=0}\end{array}\right.\right.}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Substitute back into the homogeneous equation: }} \\ {\qquad(1-t) y^{\prime \prime}+t y^{\prime}-y=0} \\ {\text { Verfified that } y_{1}(t) \text { and } y_{2}(t) \text { both satisfy the corresponding homogeneous }} \\ {\text { equation. }}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { And the complementary solution } y_{c}(t)=c_{1} e^{t}+c_{2}} \\ {\text { Now divide both sides of the original equation by } 1-t:} \\ {\qquad y^{\prime \prime}+\frac{t}{1-t} y'-\frac{1}{1-t} y=-2(t-1) e^{-t}} \\ {\qquad p(t)=\frac{t}{1-t}, g(t)=-2(t-1) e^{-t}} \\ {\qquad W\left[y_{1}, y_{2}\right](t)=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right|=(1-t) e^{t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { since the particular solution has the form: }} \\ {\qquad Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)}\end{array}
\end{equation}
\begin{equation}
\begin{aligned} u_{1}(t) &=-\int \frac{y_{2}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t \\ &=-\int \frac{t \cdot\left(-2(t-1) e^{-t}\right)}{(1-t) e^{t}} d t \\ &=-2 \int t e^{-2 t} d t \\ &=\left(t+\frac{1}{2}\right) e^{-2 t} \end{aligned}
\end{equation}
\begin{equation}
\begin{aligned} u_{2}(t) &=\int \frac{y_{1}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t \\ &=\int \frac{e^{t^{t} \cdot\left(-2(t-1) e^{-t}\right)}}{(1-t) e^{t}} d t \\ &=2 \int e^{-t} \\ &=-2 e^{-t} \end{aligned}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Therefore, }} \\ {\qquad Y(t)=\left(t+\frac{1}{2}\right) e^{-2 t} \cdot e^{t}+\left(-2 e^{-t}\right) \cdot t=\left(\frac{1}{2}-t\right) e^{-t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Hence, the general solution: }} \\ {\qquad \begin{aligned} y(t) &=y_{c}(t)+Y(t) \\ &=c_{1} e^{t}+c_{2} t+\left(\frac{1}{2}-t\right) e^{-t} \end{aligned}} \\ {\text { Therefore, the particular solution of the given nonhomogeneous equation is }} \\ {\qquad Y(t)=\left(\frac{1}{2}-t\right) e^{-t}}\end{array}
\end{equation}
6
\begin{equation}
\begin{array}{l}{y^{\prime \prime}+2 y^{\prime}+2 y=0} \\ {\text { The characteristic equation is, } r^{2}+2 r+2=0} \\ {r=\frac{-2 \pm \sqrt{4-8}}{2}} \\ {r=-1 \pm i}\end{array}
\end{equation}
\begin{equation}
\begin{array}{c}{\text { If the roots of charateristic equations are complex numbers } \lambda \pm i \mu, \mu \neq 0} \\ {\text { then the general equation is } y=c_{1}e^{\lambda t} \cos (\mu t)+c_{2}e^{\lambda t} \sin (\mu t)} \\ {\text { Compare } r=-1 \pm i \text { with } \lambda \pm i \mu} \\ {\text { we have } \lambda=-1 \quad \mu=1} \\ {\text { Subsitute the value of } \lambda=-1 \quad \mu=1} \\ {y=c_{1} e^{-t} \cos (t)+c_{2} e^{-t} \sin (t)}\end{array}
\end{equation}
\begin{array}{l}{y^{\prime \prime}+2 y^{\prime}+2 y=0} \\ {\text { The characteristic equation is, } r^{2}+2 r+2=0} \\ {r=\frac{-2 \pm \sqrt{4-8}}{2}} \\ {r=-1 \pm i}\end{array}
\end{equation}
\begin{equation}
\begin{array}{c}{\text { If the roots of charateristic equations are complex numbers } \lambda \pm i \mu, \mu \neq 0} \\ {\text { then the general equation is } y=c_{1}e^{\lambda t} \cos (\mu t)+c_{2}e^{\lambda t} \sin (\mu t)} \\ {\text { Compare } r=-1 \pm i \text { with } \lambda \pm i \mu} \\ {\text { we have } \lambda=-1 \quad \mu=1} \\ {\text { Subsitute the value of } \lambda=-1 \quad \mu=1} \\ {y=c_{1} e^{-t} \cos (t)+c_{2} e^{-t} \sin (t)}\end{array}
\end{equation}
7
Quiz-3 / TUT 0401 Quiz 3
« on: October 11, 2019, 02:08:56 PM »
\begin{equation}
y^{\prime \prime}+5 y^{\prime}+3 y=0, y(0)=1, y^{\prime}(0)=0
\end{equation}
\begin{array}{c}{\text { We assume that } y=e^{r t} \text { , and then it follows that } r \text { must }} \\ {\text { be a root of characteristic equation }} \\ {r^{2}+5 r+3=0}\end{array}
\begin{array}{l}{\text { We use the quadratic formula which is }} \\ {\qquad r=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}}\end{array}
\begin{equation}
\text { Hence, }\left\{\begin{array}{l}{r_{1}=\frac{-5+\sqrt{13}}{2}} \\ {r_{2}=\frac{-5-\sqrt{13}}{2}}\end{array}\right.
\end{equation}
\begin{equation}
\begin{array}{c}{\text { Since the general solution has the form of }} \\ {y=c_{1} e^{r_{1} t}+c_{2} e^{r_{2} t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{c}{\text { Therefore, the general solution of the given equation is }} \\ {\qquad y=c_{1} e^{\frac{\sqrt{13}-5}{2} t}+c_{2} e^{\frac{-5-\sqrt{13}}{2} t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { since } y(0)=1, y^{\prime}(0)=0} \\ {\qquad\left\{\begin{array}{l}{c_{1}+c_{2}=1} \\ {\frac{-5+\sqrt{13}}{2} c_{1}+\left(\frac{-5-\sqrt{13}}{2}\right) c_{2}=0}\end{array} \Rightarrow\left\{\begin{array}{l}{c_{1}=\frac{13+5 \sqrt{13}}{26}} \\ {c_{2}=\frac{13-5 \sqrt{13}}{26}}\end{array}\right.\right.}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Therefore, the solution of the inital value problem is }} \\ {\qquad y=\frac{13+5 \sqrt{13}}{26} e^{\frac{-5+\sqrt{13}}{2} t}+\frac{13-5 \sqrt{13}}{26} e^{\frac{-5-\sqrt{13}}{2}} t}\end{array}
\end{equation}
y^{\prime \prime}+5 y^{\prime}+3 y=0, y(0)=1, y^{\prime}(0)=0
\end{equation}
\begin{array}{c}{\text { We assume that } y=e^{r t} \text { , and then it follows that } r \text { must }} \\ {\text { be a root of characteristic equation }} \\ {r^{2}+5 r+3=0}\end{array}
\begin{array}{l}{\text { We use the quadratic formula which is }} \\ {\qquad r=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}}\end{array}
\begin{equation}
\text { Hence, }\left\{\begin{array}{l}{r_{1}=\frac{-5+\sqrt{13}}{2}} \\ {r_{2}=\frac{-5-\sqrt{13}}{2}}\end{array}\right.
\end{equation}
\begin{equation}
\begin{array}{c}{\text { Since the general solution has the form of }} \\ {y=c_{1} e^{r_{1} t}+c_{2} e^{r_{2} t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{c}{\text { Therefore, the general solution of the given equation is }} \\ {\qquad y=c_{1} e^{\frac{\sqrt{13}-5}{2} t}+c_{2} e^{\frac{-5-\sqrt{13}}{2} t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { since } y(0)=1, y^{\prime}(0)=0} \\ {\qquad\left\{\begin{array}{l}{c_{1}+c_{2}=1} \\ {\frac{-5+\sqrt{13}}{2} c_{1}+\left(\frac{-5-\sqrt{13}}{2}\right) c_{2}=0}\end{array} \Rightarrow\left\{\begin{array}{l}{c_{1}=\frac{13+5 \sqrt{13}}{26}} \\ {c_{2}=\frac{13-5 \sqrt{13}}{26}}\end{array}\right.\right.}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Therefore, the solution of the inital value problem is }} \\ {\qquad y=\frac{13+5 \sqrt{13}}{26} e^{\frac{-5+\sqrt{13}}{2} t}+\frac{13-5 \sqrt{13}}{26} e^{\frac{-5-\sqrt{13}}{2}} t}\end{array}
\end{equation}
8
Quiz-2 / Quiz 2 TUT0401
« on: October 04, 2019, 02:00:33 PM »
Here is the quiz question and solution
\begin{equation}
\left(3 x^{2} y+2 x y+y^{3}\right)+\left(x^{2}+y^{2}\right) y^{\prime}=0
\end{equation}
Set
\begin{equation}
M(x, y)=3 x^{2} y+2 x y+y^{3} \quad N(x, y)=x^{2}+y^{2}
\end{equation}
Then
\begin{eqnarray}
M_y(x, y)&=&\frac{\partial}{\partial y}\left(3 x^{2} y+2 x y+y^{3}\right)=3 x^{2}+2 x+3 y^{2} \\
N_x(x, y)&=&\frac{\partial}{\partial x}\left(x^{2}+y^{2}\right)=2 x
\end{eqnarray}
Since $M_y$ is not equal to $N_x$, the given differential equation is not exact.
\begin{eqnarray}
R&=&\frac{M y-N x}{N}=\frac{3 x^{2}+2 x+3 y^{2}-2 x}{x^{2}+y^{2}}=\frac{3\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)}=3 \\
\mu(x, y)&=&e^{\int R d x}=e^{\int 3 d x}=e^{3 x}
\end{eqnarray}
The given integrating factor is $\mu(x, y)=e^{3 x}$, multiplying both sides of the given equation with
\begin{equation}
e^{3 x}\left(3 x^{2} y+2 x y+y^{3}\right) d x+e^{3 x}\left(x^{2}+y^{2}\right) d y=0
\end{equation}
We have
\begin{eqnarray}
M(x, y)&=&3 x^{2} y e^{3 x}+2 x y e^{3 x}+y^{3} e^{3 x}\\
N(x, y)&=&e^{3 x} x^{2}+e^{3 x} y^{2}
\end{eqnarray}
Thus
\begin{eqnarray}
\begin{aligned}
M_y(x, y)&=\frac{\partial}{\partial y}\left(3 x^{2} y \cdot e^{3 x}+2 x y e^{3 x}+y^{3} e^{3 x}\right)=2 x e^{3 x}+3 x^{2} e^{3 x}+3 y^{2} e^{3 x},\\
N_{x}(x, y) &=\frac{\partial}{\partial x}\left(e^{3 x} x^{2}+e^{3 x} y^{2}\right) \\
&=2 x e^{3 x}+3 x^{2} e^{3 x}+3 y^{2} e^{3 x}.
\end{aligned}
\end{eqnarray}
Since $M_y=N_x$, the differential equation is exact now.
There exist
\begin{eqnarray}
\begin{aligned}
\psi(x, y) \text { s.t } \psi(y) &=x^{2} e^{3 x}+y^{2} e^{3 x} \\ \psi(x, y) &=\int x^{2} e^{3 x}+y^{2} e^{3 x} d y \\ &=x^{2} y e^{3 x}+\frac{1}{3} y^{3} e^{3 x}+h(x) \end{aligned}
\end{eqnarray}
\begin{equation}
\psi_x=2x e^{3x} y+3x^2 e^{3x}y+y^3 e^{3x}+h'(x)=3x^2y e^{3x}+2y e^{3x}x+y^3e^{3x}
\end{equation}
gives $h'(x)=0$, which implies $h(x)=C$.
Substitute $h(x)=C$ in equation
\begin{eqnarray}
\begin{aligned} \psi(x, y)&=x^{2} y e^{3 x}+\frac{1}{3} y^{3} e^{3 x}+h(x) \\
\psi(x, y) &=x^{2}y e^{3 x} +\frac{1}{3} y^{3} e^{3 x}+C \\ &=x^{2} ye^{3 x} +\frac{1}{3} y^{3} e^{3 x} \end{aligned}
\end{eqnarray}
Then the solution of differential equation is
\begin{eqnarray}
\begin{array}{c}{} {\qquad \begin{array}{r}{x^{2} e^{3 x} y+\frac{1}{3} y^{3} e^{3 x}=k} \\ {3 x^{2} e^{3 x} y+y^{3} e^{3 x}=3 k} \\ \end{array}}\end{array}
\end{eqnarray}
We have
\begin{equation}
{\left(3 x^{2} y+y^{3}\right) e^{3 x}=C \quad (3 k=C)}
\end{equation}
Hence, the required solution of the differential equation is
\begin{eqnarray}
\left(3x^{2} y+y^{3}\right) e^{3 x}=C
\end{eqnarray}
\begin{equation}
\left(3 x^{2} y+2 x y+y^{3}\right)+\left(x^{2}+y^{2}\right) y^{\prime}=0
\end{equation}
Set
\begin{equation}
M(x, y)=3 x^{2} y+2 x y+y^{3} \quad N(x, y)=x^{2}+y^{2}
\end{equation}
Then
\begin{eqnarray}
M_y(x, y)&=&\frac{\partial}{\partial y}\left(3 x^{2} y+2 x y+y^{3}\right)=3 x^{2}+2 x+3 y^{2} \\
N_x(x, y)&=&\frac{\partial}{\partial x}\left(x^{2}+y^{2}\right)=2 x
\end{eqnarray}
Since $M_y$ is not equal to $N_x$, the given differential equation is not exact.
\begin{eqnarray}
R&=&\frac{M y-N x}{N}=\frac{3 x^{2}+2 x+3 y^{2}-2 x}{x^{2}+y^{2}}=\frac{3\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)}=3 \\
\mu(x, y)&=&e^{\int R d x}=e^{\int 3 d x}=e^{3 x}
\end{eqnarray}
The given integrating factor is $\mu(x, y)=e^{3 x}$, multiplying both sides of the given equation with
\begin{equation}
e^{3 x}\left(3 x^{2} y+2 x y+y^{3}\right) d x+e^{3 x}\left(x^{2}+y^{2}\right) d y=0
\end{equation}
We have
\begin{eqnarray}
M(x, y)&=&3 x^{2} y e^{3 x}+2 x y e^{3 x}+y^{3} e^{3 x}\\
N(x, y)&=&e^{3 x} x^{2}+e^{3 x} y^{2}
\end{eqnarray}
Thus
\begin{eqnarray}
\begin{aligned}
M_y(x, y)&=\frac{\partial}{\partial y}\left(3 x^{2} y \cdot e^{3 x}+2 x y e^{3 x}+y^{3} e^{3 x}\right)=2 x e^{3 x}+3 x^{2} e^{3 x}+3 y^{2} e^{3 x},\\
N_{x}(x, y) &=\frac{\partial}{\partial x}\left(e^{3 x} x^{2}+e^{3 x} y^{2}\right) \\
&=2 x e^{3 x}+3 x^{2} e^{3 x}+3 y^{2} e^{3 x}.
\end{aligned}
\end{eqnarray}
Since $M_y=N_x$, the differential equation is exact now.
There exist
\begin{eqnarray}
\begin{aligned}
\psi(x, y) \text { s.t } \psi(y) &=x^{2} e^{3 x}+y^{2} e^{3 x} \\ \psi(x, y) &=\int x^{2} e^{3 x}+y^{2} e^{3 x} d y \\ &=x^{2} y e^{3 x}+\frac{1}{3} y^{3} e^{3 x}+h(x) \end{aligned}
\end{eqnarray}
\begin{equation}
\psi_x=2x e^{3x} y+3x^2 e^{3x}y+y^3 e^{3x}+h'(x)=3x^2y e^{3x}+2y e^{3x}x+y^3e^{3x}
\end{equation}
gives $h'(x)=0$, which implies $h(x)=C$.
Substitute $h(x)=C$ in equation
\begin{eqnarray}
\begin{aligned} \psi(x, y)&=x^{2} y e^{3 x}+\frac{1}{3} y^{3} e^{3 x}+h(x) \\
\psi(x, y) &=x^{2}y e^{3 x} +\frac{1}{3} y^{3} e^{3 x}+C \\ &=x^{2} ye^{3 x} +\frac{1}{3} y^{3} e^{3 x} \end{aligned}
\end{eqnarray}
Then the solution of differential equation is
\begin{eqnarray}
\begin{array}{c}{} {\qquad \begin{array}{r}{x^{2} e^{3 x} y+\frac{1}{3} y^{3} e^{3 x}=k} \\ {3 x^{2} e^{3 x} y+y^{3} e^{3 x}=3 k} \\ \end{array}}\end{array}
\end{eqnarray}
We have
\begin{equation}
{\left(3 x^{2} y+y^{3}\right) e^{3 x}=C \quad (3 k=C)}
\end{equation}
Hence, the required solution of the differential equation is
\begin{eqnarray}
\left(3x^{2} y+y^{3}\right) e^{3 x}=C
\end{eqnarray}
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