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### Messages - Tiantian Yu

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##### Quiz-4 / TUT0303 Quiz4
« on: October 18, 2019, 02:00:00 PM »
Find the general solution of the given differential equation: y''+4y'+6.25y=0

solution: The auxiliary equation is \begin{equation*}r^2+4r+6.25=0\end{equation*}
By the quadratic formula, the roots are \begin{equation*}r=\frac{-4\pm\sqrt{4^2 - 4(6.25)}}{2}=-2\pm\frac{3}{2}i.\end{equation*}
If the roots of the auxiliary equation are the complex numbers 􏲬\begin{equation*}r=\alpha\pm i\beta\end{equation*} then the general solution of ay''+by'+cy=0 is\begin{equation*} y=e^{\alpha t} (C_1 cos\beta t +C_2sin\beta t)\end{equation*}
since􏲬\begin{equation*}\alpha=-2,\beta=\frac{3}{2}\end{equation*}
the general solution of the differential equation is \begin{equation*} y=e^{-2 t} (C_1 cos\frac{3}{2} t +C_2sin\frac{3}{2} t)\end{equation*}

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##### Quiz-3 / TUT0303 Quiz3
« on: October 11, 2019, 02:00:39 PM »
Verify that the functions y1 and y2 are solutions of the given differential equation. Do they constitute a fundamental set of solutions?
y'' + 4y = 0;  y1(t)=cos(2t), y2(t)=sin(2t)

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solution: firstly verifying that the functions y1 and y2 are solutions of the given differential equation
verify y1(t)=cos(2t):
Left-hand side=y'' + 4y=(cos(2t))''+4(cos(2t))
=(-2sin(2t))'+4(cos(2t))
=-4cos(2t)+4(cos(2t))
=0=Right-hand side

verify y2(t)=sin(2t):
Left-hand side=y'' + 4y=((sin(2t))''+4(sin(2t))
=((2cos(2t))'+4(sin(2t))
=(-4sin(2t)+4(sin(2t))
=0=Right-hand side
Thus the functions y1 and y2 are solutions of the given differential equation.

Now we are checking if the given functions constitute a fundamental set of solutions
We calculate the Wronskian of y1 and y2:
\begin{equation*}
W=\begin{bmatrix}
cos(2t)  &  sin(2t)\\
-2sin(2t)  &   2cos(2t)
\end{bmatrix}\end{equation*}
\begin{equation*}=cos(2t)(2cos(2t))-[sin(2t)(-2sin(2t)]\end{equation*}
\begin{equation*}=2cos^2(2t)-(-2sin^2(2t))\end{equation*}
\begin{equation*}=2(cos^2(2t)+sin^2(2t))\end{equation*}
\begin{equation*}=2\end{equation*}
Since the Wronskian of y1 and y2 is not zero, which means they are linearly independent, the functions y1 and y2 constitute a fundamental set of solutions

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##### Quiz-2 / TUT0303 Quiz2
« on: October 04, 2019, 05:57:49 PM »
find the value of b for which the given equation is exact, and then solve it using that value of b.
\begin{equation*}
(xy^2+bx^2y)+(x+y)x^2y'=0
\end{equation*}
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Solution: By calculating My and Nx , we find that
\begin{equation*} M_y(x,y)= \frac{d}{dy} (xy^2+bx^2y) = 2xy+bx^2 \end{equation*} \begin{equation*} N_x(x,y)= \frac{d}{dx} ((x+y)x^2) =  \frac{d}{dx} (x^3+yx^2)= 3x^2+2xy \end{equation*}
since we know that the given equation is exact, and it is exact when My=Nx
\begin{equation*}2xy+bx^2=3x^2+2xy\end{equation*}
thus we get b=3

there is a ψ(x, y) such that
\begin{equation*}ψ_x(x,y)=xy^2+3x^2y \end{equation*} \begin{equation*}ψ_y(x,y)=(x+y)x^2=x^3+yx^2\end{equation*}
Integrating the first of these equations with respect to x, we obtain \begin{equation*}ψ (x , y )=\int(xy^2+3x^2y) dx=\frac{1}{2} x^2y^2+x^3y+h(y) \end{equation*}
Then computing ψy from the equation above and setting ψy = N gives
\begin{equation*}ψ_y(x,y) =\frac{d}{dy} (\frac{1}{2} x^2y^2+x^3y+h(y))=x^2y+x^3+h'(y)=x^3+yx^2\end{equation*}
Thus h′(y)=0, and h(y)=C
Substituting for h(y) in ψ(x, y) gives \begin{equation*}ψ(x,y) =\frac{1}{2} x^2y^2+x^3y+C \end{equation*}
Hence the solution is given implicitly by \begin{equation*}C=\frac{1}{2} x^2y^2+x^3y=x^2y^2+2x^3y\end{equation*}

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##### Quiz-1 / TUT0303 Quiz1
« on: September 27, 2019, 07:03:14 PM »
Find the general solution of
\begin{equation*}ty'-y=t^{2}e^{-t}
\end{equation*}
and use it to determine how solutions behave as t → ∞.

put the differential equation into standard form: \begin{equation*}y'-\frac{y}{t}=\frac{t^{2}e^{-t}}{t}\end{equation*}
then \begin{equation*}p(t)=-\frac{1}{t}\end{equation*}
the integrating factor is \begin{equation*}μ(t)= e^{\int-\frac{1}{t}dt}=e^{-ln|t|} =\frac{1}{|t|}\end{equation*}
since t>0,we take\begin{equation*}\frac{1}{t}\end{equation*}
\begin{equation*}\frac{d}{dx}(\frac{1}{t}y)=\frac{t^{2}e^{-t}}{t}\frac{1}{t}\end{equation*}
integrating both sides, we have\begin{equation*}\frac{1}{t}y=\int\frac{t^{2}e^{-t}}{t^{2}}dt=-e^{-t}+C\end{equation*}
\begin{equation*}y=-te^{-t}+Ct\end{equation*}
the behaviour of the solution is determined by the term Ct
if c=0,\begin{equation*}\lim_{t\to\infty} -te^{-t}=-\lim_{t\to\infty} te^{-t}=0\end{equation*}
if c is positive,\begin{equation*}\lim_{t\to\infty} -te^{-t}+Ct=\lim_{t\to\infty} Ct=\infty\end{equation*}
if c is negative,\begin{equation*}\lim_{t\to\infty} -te^{-t}-Ct=-\lim_{t\to\infty} Ct=-\infty\end{equation*}
Thus the general solution of the given differential equation is\begin{equation*}y=-te^{-t}+Ct\end{equation*}and y=∞,-∞, or 0 as t->∞

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