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Messages - Monika Dydynski

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Thanksgiving Bonus / Re: Thanksgiving bonus 3
« on: October 05, 2018, 07:12:37 PM »
Find a second order equation with the fundamental system of solutions $\lbrace y_1(x),y_2(x) \rbrace  =\lbrace x^3,e^x\rbrace$

$$W(x^3,e^x)=\left|\begin{matrix}x^3&e^x\\ 3x^2&e^x\end{matrix}\right|=x^3e^x-3x^2e^x\ne0$$

$$W(y,x^3,e^x)=\left|\begin{matrix}y&x^3&e^x\\ y'&3x^2&e^x\\y''&6x&e^x\end{matrix}\right|=y(3x^2e^x-6xe^x)-y'(x^3e^x-6xe^x)+y''(x^3e^x-3x^2e^x)=0$$

Since $x^3e^x-3x^2e^x\ne0$,

$$y''-{(x^3e^x-6xe^x) \over (x^3e^x-3x^2e^x)}y'+{(3x^2e^x-6xe^x) \over (x^3e^x-3x^2e^x)}y=0$$

Thus, a second order equation with the fundamental system of solutions $\lbrace y_1(x),y_2(x) \rbrace  =\lbrace x^3,e^x\rbrace$ is

$$y''-{(x^2-6) \over (x^2-3x)}y'+{(3x-6) \over (x^2-3x)}y=0$$

Quiz-2 / Re: Q2 TUT 0101, TUT 0501 and TUT 0801
« on: October 05, 2018, 05:29:40 PM »
Find an integrating factor and solve the given equation.
$$1+\left({x \over y}−\sin(y)\right)y′=0.\tag{1}$$
Let $M(x,y)=1$ and $N(x,y)={x \over y}−\sin(y)$.
Then, $M_y={\partial \over \partial y}M(x,y)=0$ and $N_x={\partial \over \partial x}N(x,y)={1 \over y}$.
Notice that $M_y =0\ne{1 \over y} =N_x$, which implies that the given equation is not exact.
We are looking for an integrating factor $\mu(x,y)$ such that after multiplying $(1)$ by $\mu$, the equation becomes exact.
That is, $(\mu M)_y=(\mu N)_x$.
 $${\partial \mu \over \partial y }={\partial \over \partial x}\left[\mu\left({x \over y}-\sin(y)\right)\right]$$
$${\partial \mu \over \partial y }={\partial \mu \over \partial x }\left({x \over y}-\sin(y)\right)+\mu {1 \over y}$$
Suppose that $\mu$ is a function of only $y$, we get
$${d \mu \over dy}=\mu {1 \over y}$$
$$\int{1 \over \mu}d \mu=\int{1 \over y}dy $$
Multiplying $(1)$ by the integrating factor $\mu=y$, we get an exact equation,
$$y+(x-y \sin(y))y'=0\tag{2}$$
Since $(2)$ is exact, there exists a solution $\phi (x,y)=C$ such that
$\phi_y(x,y)=x-y \sin(y)\tag{4}$
Integrating $(3)$ with respect to $x$, we get
for some function $g$ of $y$.
Differentiating $(5)$ with respect to $y$, we get
Equating $(4)$ with $(6)$, we have
$$\int -y\sin(y)dy=\int g'(y)dy$$
$$g(y)=-\int y\sin(y)dy$$
Integration by parts gives,
Substituting $(7)$ into $(5)$,
Thus, the solutions of the differential equation are given implicitly by

Quiz-1 / Re: Readme before posting
« on: September 28, 2018, 06:07:10 PM »
If someone has posted a scanned solution, are we allowed to post a typed solution since typed solutions are preferred? Or does the "no posts after a perfect solution" rule still apply?

Also Mengmeng, i think you took the professor's post too literally. I believe when he writes "Post solutions to Quiz-1 here", he means that we should reply to Quiz 1 questions in the Quiz 1 Board, but directly in response to the question. NOT in response to this post lol

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